Atomic Structure MCQs with Answers – Part 7 (NEET Chemistry)

Explanation: \(\textbf{Express the first photoelectron kinetic energy:}\) Einstein's photoelectric equation gives \[ K_{\max}=hf-\phi \] For incident frequency \(f\), \[ K_1=hf-\phi \] For incident frequency \(3f\), \[ K_3=3hf-\phi \] \(\textbf{Use the measured kinetic-energy ratio:}\) The problem states \[ K_3=5K_1 \] Substituting, \[ 3hf-\phi=5(hf-\phi) \] Expanding, \[ 3hf-\phi=5hf-5\phi \] Rearranging, \[ 4\phi=2hf \] \[ \phi=\frac{hf}{2} \] \(\textbf{Recover the threshold frequency:}\) The work function is related to the threshold frequency by \[ \phi=hf_0 \] Therefore, \[ hf_0=\frac{hf}{2} \] \[ f_0=\frac f2 \] This is consistent with the stipulated fact that both original frequencies exceed threshold. \(\textbf{Calculate the original maximum kinetic energy:}\) \[ K_1=hf-\frac{hf}{2} \] \[ K_1=\frac{hf}{2} \] \(\textbf{Apply the new frequency }2f\textbf{:}\) \[ K_2=h(2f)-\phi \] Substituting the inferred work function, \[ K_2=2hf-\frac{hf}{2} \] \[ K_2=\frac{3hf}{2} \] \(\textbf{Compare the two kinetic energies:}\) \[ \frac{K_2}{K_1} =\frac{3hf/2}{hf/2} \] \[ \frac{K_2}{K_1}=3 \] \(\textbf{Check against the original measurement:}\) At frequency \(3f\), \[ K_3=3hf-\frac{hf}{2}=\frac{5hf}{2} \] This gives \[ \frac{K_3}{K_1}=5 \] as observed. \(\textbf{Interpret the non-proportional kinetic-energy changes:}\) Photoelectron kinetic energy is linear in frequency but does not pass through the origin because the work function must first be supplied. Tripling incident frequency therefore need not triple maximum electron kinetic energy. The measured ratio instead reveals the work function relative to \(hf\). That information allows the response at an intermediate frequency to be predicted. \(\textbf{Final answer:}\) \[ \boxed{f_0=\frac f2,\qquad\frac{K_2}{K_1}=3} \]

Explanation: \(\textbf{Convert all observations to a common wavelength scale:}\) For radiation P, \[ \lambda_P=300\,\mathrm{nm} \] For Q, the vacuum wave relation is \[ c=f_Q\lambda_Q \] Therefore, \[ \lambda_Q=\frac{c}{f_Q} \] \[ \lambda_Q=\frac{3.00\times10^8}{1.50\times10^{15}}\,\mathrm m \] \[ \lambda_Q=2.00\times10^{-7}\,\mathrm m \] \[ \lambda_Q=200\,\mathrm{nm} \] For R, wavenumber is reciprocal wavelength: \[ \tilde{\nu}_R=\frac1{\lambda_R} \] Thus, \[ \lambda_R=\frac1{4.00\times10^6}\,\mathrm m \] \[ \lambda_R=2.50\times10^{-7}\,\mathrm m \] \[ \lambda_R=250\,\mathrm{nm} \] \(\textbf{Rank the energy of one photon:}\) Photon energy varies inversely with vacuum wavelength: \[ E_\gamma=\frac{hc}{\lambda} \] The wavelengths satisfy \[ \lambda_Q\lt \lambda_R\lt \lambda_P \] Consequently, \[ E_Q\gt E_R\gt E_P \] \(\textbf{Apply the equal-power condition:}\) Optical power is photon energy delivered per unit time: \[ \mathcal P=\dot N E_\gamma \] The three incident powers are equal. Therefore, \[ \dot N=\frac{\mathcal P}{E_\gamma} \] The photon arrival rate is inversely proportional to the energy of one photon. Thus, \[ \dot N_P\gt \dot N_R\gt \dot N_Q \] \(\textbf{Interpret the two opposite rankings:}\) The highest-frequency radiation has the greatest energy per photon. However, fewer of those higher-energy photons are needed to deliver a specified amount of energy per second. Therefore, equal-power beams need not supply equal photon numbers. Frequency and wavenumber must be converted into the same physical wavelength scale before making either comparison. It also distinguishes frequency from wavenumber: they have different units even though both increase when wavelength decreases. The two rankings are opposite because photon energy and photon arrival rate respond inversely under the equal-power condition.

Explanation: \(\textbf{Apply the original minimum-uncertainty preparations:}\) Heisenberg's inequality is \[ \Delta x\,\Delta p_x\geq\frac{\hbar}{2} \] At the stipulated minimum, \[ \Delta p_x=\frac{\hbar}{2\Delta x} \] For beam P, \[ \Delta p_{x,P}=\frac{\hbar}{2d} \] For beam Q, \[ \Delta p_{x,Q}=\frac{\hbar}{2(2d)} \] \[ \Delta p_{x,Q}=\frac{\hbar}{4d} \] Thus, \[ \frac{\Delta p_{x,Q}}{\Delta p_{x,P}}=\frac12 \] \(\textbf{Compare initial fractional uncertainties:}\) Let the common initial central momentum magnitude be \(p\). For P, \[ f_P=\frac{\Delta p_{x,P}}p=\frac{\hbar}{2dp} \] For Q, \[ f_{Q,i}=\frac{\hbar}{4dp} \] Therefore, \[ f_{Q,i}=\frac{f_P}{2} \] \(\textbf{Apply the changed central momentum:}\) The device doubles Q's central momentum: \[ p_{Q,f}=2p \] The problem explicitly states that Q's absolute uncertainty remains \[ \Delta p_{x,Q,f}=\frac{\hbar}{4d} \] Its final fractional uncertainty is consequently \[ f_{Q,f}=\frac{\Delta p_{x,Q,f}}{p_{Q,f}} \] \[ f_{Q,f}=\frac{\hbar/(4d)}{2p} \] \[ f_{Q,f}=\frac{\hbar}{8dp} \] \(\textbf{Compare with the original fraction for P:}\) \[ \frac{f_{Q,f}}{f_P} =\frac{\hbar/(8dp)}{\hbar/(2dp)} \] \[ \frac{f_{Q,f}}{f_P}=\frac14 \] \(\textbf{Interpret the two independent changes:}\) Q initially has a larger position uncertainty and therefore a smaller minimum absolute momentum uncertainty. The later device changes Q's central momentum but not the width of its momentum distribution. Consequently, the fractional momentum uncertainty decreases although the absolute uncertainty remains fixed. The uncertainty principle constrains the product of the spreads, not the numerical value of the mean momentum. Doubling a beam's central momentum does not automatically require doubling its momentum uncertainty. This distinguishes a change in the average or central motion from a change in the precision of the quantum state. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Delta p_{x,Q}}{\Delta p_{x,P}}=\frac12,\qquad\frac{f_{Q,f}}{f_P}=\frac14} \]

Explanation: \(\textbf{Square the two supplied wave functions:}\) The local spatial probability density is the squared magnitude of the wave function. For \(1s\), \[ \rho_{1s}(r)=\frac1{\pi a_0^3}e^{-2r/a_0} \] For \(2s\), \[ \rho_{2s}(r)=\frac1{32\pi a_0^3} \left(2-\frac r{a_0}\right)^2e^{-r/a_0} \] \(\textbf{Form a common-radius density ratio:}\) Dividing the two expressions, \[ \frac{\rho_{2s}(r)}{\rho_{1s}(r)} =\frac1{32}\left(2-\frac r{a_0}\right)^2e^{r/a_0} \] This relation includes both the polynomial nodal factor and the different exponential radial dependences. \(\textbf{Evaluate the ratio at the nucleus:}\) At \(r=0\), \[ \left(\frac{\rho_{2s}}{\rho_{1s}}\right)_0 =\frac1{32}(2)^2e^0 \] \[ \left(\frac{\rho_{2s}}{\rho_{1s}}\right)_0 =\frac4{32} \] \[ \left(\frac{\rho_{2s}}{\rho_{1s}}\right)_0=\frac18 \] Therefore, the \(1s\) local density is eight times the \(2s\) local density at the nucleus. \(\textbf{Evaluate the ratio at }r=4a_0\textbf{:}\) Substituting \(r/a_0=4\), \[ \left(\frac{\rho_{2s}}{\rho_{1s}}\right)_{4a_0} =\frac1{32}(2-4)^2e^4 \] \[ \left(\frac{\rho_{2s}}{\rho_{1s}}\right)_{4a_0} =\frac{e^4}{8} \] Using the supplied exponential value, \[ \left(\frac{\rho_{2s}}{\rho_{1s}}\right)_{4a_0} \approx\frac{54.6}{8} \] \[ \left(\frac{\rho_{2s}}{\rho_{1s}}\right)_{4a_0}\approx6.83 \] \(\textbf{Check the radial-node condition:}\) The \(2s\) wave function vanishes where \[ 2-\frac r{a_0}=0 \] Thus, its radial node occurs at \[ r=2a_0 \] At this radius the \(2s\) density is zero, but the \(1s\) density is positive. \(\textbf{Interpret the spatial-density reversal:}\) The \(1s\) density is more concentrated near the nucleus. The \(2s\) wave function has a radial node and a more extended outer region. Its probability density can consequently exceed the \(1s\) density at a sufficiently large radius. Neither density is negative, even though the real \(2s\) wave function changes sign across its node. The local-density comparison also does not directly give the total probability outside a spherical boundary, which requires integration over volume. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{\rho_{2s}}{\rho_{1s}}\right)_0=\frac18,\qquad \left(\frac{\rho_{2s}}{\rho_{1s}}\right)_{4a_0}\approx6.83} \]

Explanation: The normalisation condition assigns total spatial probability one to an orbital state. For the original wave function, \[ \int|\Psi|^2\,dV=1 \] Multiplication of a wave function by a constant changes the integral of its squared magnitude unless the constant has magnitude one. For P, the transformed wave function is \(-\Psi\). Its squared magnitude is \[ |-\Psi|^2=|\Psi|^2 \] Therefore, P remains normalised and gives exactly the same spatial probability density. The overall sign change is not an observable change in electron-position probability. For Q, the transformed wave function is \(2\Psi\). Its squared magnitude is four times that of the original wave function. Consequently, its integral over all space is four rather than one. It is therefore not normalised as written. For R, the transformed wave function is \(\Psi/2\). Its squared magnitude is one-fourth of the original density. Its probability integral is consequently one-fourth rather than one. Neither Q nor R can be used directly as a normalised orbital wave function in its stated form. However, both retain the original wave function's spatial dependence and nodal locations up to an overall nonzero scale. After renormalisation, they describe the same spatial probability distribution as \(\Psi\). The investigator is correct that multiplication by a nonzero constant need not change the physical spatial state, but incorrect that every such rescaled expression is already normalised. The distinction is between an unchanged state after appropriate normalisation and an unchanged numerical probability density before normalisation.

Explanation: The two selected positions have opposite Cartesian coordinates. However, their distances from the nucleus are equal: \[ r_A=\sqrt{a^2+b^2+c^2} \] \[ r_B=\sqrt{(-a)^2+(-b)^2+(-c)^2}=r_A \] Therefore, the radial factors have the same values at both positions. For P, the wave function is proportional to the \(x\) coordinate. At A, \[ \Psi_P(A)=af(r_A) \] At B, \[ \Psi_P(B)=-af(r_A) \] Thus, P changes sign under inversion. Its squared magnitude does not change: \[ |\Psi_P(B)|^2=|\Psi_P(A)|^2 \] For Q, the angular factor is the product \(xy\). At A, \[ \Psi_Q(A)=abg(r_A) \] At B, \[ \Psi_Q(B)=(-a)(-b)g(r_A) \] \[ \Psi_Q(B)=abg(r_A) \] Therefore, Q retains its sign under inversion. Its probability density is also unchanged. The different sign behaviours arise from the different angular factors of the two orbitals. The \(x\) factor changes sign once, whereas the \(xy\) product contains two coordinate sign changes that cancel. Neither reversal nor preservation of the wave-function sign requires unequal electron-position probabilities at the opposite points. The comparison distinguishes the mathematical phase behaviour of orbital wave functions from their observable spatial probability densities. It also illustrates why different orbital shapes can be characterised through directional relationships without treating wave-function sign as a negative probability.

Explanation: \(\textbf{Interpret the pointwise spatial evidence:}\) The \(2s\) orbital possesses one finite spherical radial node at \(r=r_0\). Its wave function and local probability density therefore vanish at every point on that nodal sphere. In particular, \[ \rho_{2s}(A)=0 \] The \(2p_z\) orbital has a different nodal geometry. At A, its wave function is explicitly stated to be nonzero. Hence, \[ \rho_{2p_z}(A)\gt 0 \] The orbitals consequently have different spatial probability distributions. \(\textbf{Determine their energies in the elementary hydrogen model:}\) For a hydrogen atom, \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] Both orbitals have principal quantum number two. Therefore, \[ E_{2s}=E_{2p}=-\frac{13.6}{4} \] \[ E_{2s}=E_{2p}=-3.40\,\mathrm{eV} \] This equality is the gross-energy degeneracy of the elementary one-electron Coulombic treatment. \(\textbf{Find the common ionisation threshold:}\) The continuum reference energy is zero. Hence, the ionisation energy from either orbital is \[ E_{\mathrm{ion}}=0-(-3.40) \] \[ E_{\mathrm{ion}}=3.40\,\mathrm{eV} \] \(\textbf{Apply the incident photon energy:}\) Each incident photon carries \[ E_\gamma=4.00\,\mathrm{eV} \] This exceeds the common threshold. The electron kinetic energy is \[ K=E_\gamma-E_{\mathrm{ion}} \] \[ K=4.00-3.40 \] \[ K=0.60\,\mathrm{eV} \] \(\textbf{Interpret the independence of spatial shape and binding energy:}\) A zero of probability density at one spatial point does not prevent the electron from occupying the complete orbital or being ionised. The nodal geometry describes the wave function's spatial structure. The ionisation threshold instead depends on the total electronic energy of the initial state. In the elementary hydrogen model, states with the same principal quantum number can have different nodal structures while retaining equal energies. Consequently, the two photoelectrons can have equal kinetic energies despite the different position probabilities measured at A. \(\textbf{Final answer:}\) \[ E_{\mathrm{ion},2s}=E_{\mathrm{ion},2p}=3.40\,\mathrm{eV},\qquad K_{2s}=K_{2p}=0.60\,\mathrm{eV} \]

Explanation: A stationary atomic state can possess a characteristic electronic energy without assigning a definite classical trajectory to its electron. This is a central feature of the quantum mechanical description. The spatial orbital is represented mathematically by a wave function. Its squared magnitude gives the electron's spatial probability density. Therefore, Description P correctly combines a characteristic electronic energy with a probabilistic account of electron position. Description Q is incorrect because definite electronic energy does not remove the position-momentum uncertainty principle. A stationary state is not equivalent to an electron travelling around a precisely defined circular path with exactly known position and momentum. The uncertainty relation continues to constrain the simultaneous precision of corresponding position and momentum components. Description R is also incorrect. A nonzero or even maximum local probability density at the nucleus does not mean that the electron is permanently located there. The \(1s\) electron has a spatial probability distribution extending over a finite region around the nucleus and beyond any customary boundary surface. An electron-position measurement may yield different positions in repeated preparations. The permitted electronic energies determine the energy changes possible between bound states. For an ideal one-photon bound-bound absorption process, the photon energy must correspond to an allowed electronic energy difference. This requirement is independent of assigning the electron a classical trajectory. Thus, the probabilistic orbital description and energy quantisation are complementary features of quantum mechanics. A spatial density maximum does not itself set the absorbed photon energy; the difference between the initial and final electronic energies does. Description P is the only one consistent with both the spatial and spectroscopic aspects of the model.

Explanation: \(\textbf{Compare the radial wave-function amplitudes:}\) The supplied expression is \[ \Psi(r)=Ae^{-r/a} \] At the first radius, \[ \Psi(r_1)=Ae^{-\ln2} \] \[ \Psi(r_1)=\frac A2 \] At the second radius, \[ \Psi(r_2)=Ae^{-2\ln2} \] \[ \Psi(r_2)=\frac A4 \] Therefore, \[ \frac{\Psi(r_2)}{\Psi(r_1)}=\frac{A/4}{A/2} \] \[ \frac{\Psi(r_2)}{\Psi(r_1)}=\frac12 \] \(\textbf{Convert amplitudes to local probability densities:}\) The local spatial probability density is \[ \rho(r)=|\Psi(r)|^2 \] For the two radii, \[ \rho(r_1)=\frac{A^2}{4} \] \[ \rho(r_2)=\frac{A^2}{16} \] Thus, \[ \frac{\rho(r_2)}{\rho(r_1)}=\frac14 \] \(\textbf{Include the geometry of the spherical shells:}\) The probability associated with a thin complete spherical shell is \[ \Delta P\approx4\pi r^2\rho(r)\Delta r \] The shells have equal thickness, so \[ \frac{P_2}{P_1}=\frac{r_2^2}{r_1^2}\frac{\rho(r_2)}{\rho(r_1)} \] Since \(r_2=2r_1\), \[ \frac{P_2}{P_1}=2^2\left(\frac14\right) \] \[ \frac{P_2}{P_1}=1 \] \(\textbf{Interpret the three comparisons:}\) The wave-function amplitude decreases by a factor of two. Its squared magnitude consequently decreases by a factor of four. However, the second spherical shell has four times the geometrical area of the first. For equal small thicknesses, the area increase exactly compensates for the local-density decrease. Therefore, the two shells have equal approximate probabilities despite different wave-function amplitudes and local densities. The result illustrates why a graph of \(\Psi^2\) against radial distance must not be interpreted directly as the probability in equal-thickness spherical shells. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Psi(r_2)}{\Psi(r_1)}=\frac12,\qquad\frac{\rho(r_2)}{\rho(r_1)}=\frac14,\qquad\frac{P_2}{P_1}=1} \]

Explanation: \(\textbf{Interpret the orbital boundary:}\) The stipulated spherical boundary encloses \(90\%\) of the total electron-position probability. Therefore, \[ P_{\mathrm{inside}}=0.90 \] The probability outside is \[ P_{\mathrm{outside}}=1-0.90 \] \[ P_{\mathrm{outside}}=0.10 \] The boundary is a chosen probability surface, not a wall beyond which the electron cannot be detected. \(\textbf{Distribute the enclosed probability among octants:}\) A spherically symmetric orbital has no preferred spatial direction. The eight octants defined by the coordinate planes consequently contain equal portions of the probability enclosed by the sphere. Thus, \[ P_{\mathrm{one\ octant}}=\frac{0.90}{8} \] \[ P_{\mathrm{one\ octant}}=0.1125 \] \[ P_{\mathrm{one\ octant}}=11.25\% \] \(\textbf{Examine the probability beyond the selected boundary:}\) The total probability outside the sphere is \(10\%\). Spherical symmetry divides this probability equally between the positive-\(z\) and negative-\(z\) half-spaces. Therefore, \[ P_{\mathrm{outside},z\gt 0}=\frac{0.10}{2} \] \[ P_{\mathrm{outside},z\gt 0}=0.05 \] \[ P_{\mathrm{outside},z\gt 0}=5\% \] \(\textbf{Calculate the probability in the complete upper half-space:}\) Half of the \(90\%\) enclosed probability lies above the \(xy\)-plane. Hence, \[ P_{\mathrm{inside},z\gt 0}=\frac{0.90}{2} \] \[ P_{\mathrm{inside},z\gt 0}=0.45 \] Adding the probability outside the spherical boundary, \[ P_{z\gt 0}=0.45+0.05 \] \[ P_{z\gt 0}=0.50 \] \(\textbf{Interpret the distinct sampling regions:}\) An octant inside the boundary is not one-eighth of all space in the probability-accounting sense used here; it is one-eighth of the enclosed probability. The positive-\(z\) half-space, by contrast, extends both inside and outside the selected sphere. The complete upper half-space therefore contains \(50\%\) of the total probability. These results follow from spherical symmetry and the boundary's specified enclosed probability. They do not require the detailed radial formula of the \(1s\) wave function. \(\textbf{Final answer:}\) \[ \boxed{P_{\mathrm{octant}}=11.25\%,\qquad P_{\mathrm{outside},z\gt 0}=5\%,\qquad P_{z\gt 0}=50\%} \]

Explanation: The \(2s\) orbital has one finite radial node at \(r=r_0\). At the node, its real wave function vanishes. The local spatial probability density is therefore zero on the corresponding spherical surface. The radial probability distribution \(4\pi r^2|\Psi(r)|^2\) is also zero at that radius. However, these statements concern one radial position rather than an entire finite interval. A \(2s\) wave function becomes nonzero again outside its radial node. Its sign may be opposite to the sign in the inner region, but its squared magnitude is positive wherever the wave function is nonzero. The cumulative probability \(F(r)\) includes all probability enclosed within the sphere of radius \(r\). As the sphere expands, previously included probability is not removed. The probability added between two radii is obtained from the nonnegative radial distribution over that interval. Therefore, \(F(r)\) cannot decrease with increasing radius. A node can produce a zero radial probability density at one radius without producing a finite plateau in the cumulative probability curve. Under the ordinary \(2s\) radial distribution, every finite radial interval beyond the node contains some nonzero probability. Thus, for a radius sufficiently beyond the node, \[ F(r)\gt F(r_0)=0.30 \] while normalisation requires \(F(r)\) to approach one at large radial distance. The investigator has confused a zero in the local wave-function amplitude with the absence of probability throughout a surrounding volume. The distinction is essential when interpreting \(\Psi\), \(\Psi^2\), radial probability distributions and cumulative probability graphs for the same orbital.

Explanation: \(\textbf{Obtain the spatial probability density:}\) Squaring the supplied wave function gives \[ \rho_Z(r)=|\Psi_Z(r)|^2 \] \[ \rho_Z(r)=\frac{Z^3}{\pi a_0^3}e^{-2Zr/a_0} \] \(\textbf{Compare the two species at the nucleus:}\) At \(r=0\), the exponential factor equals one. For hydrogen, \[ \rho_{\mathrm H}(0)=\frac1{\pi a_0^3} \] For \(\mathrm{He^+}\), \[ \rho_{\mathrm{He^+}}(0)=\frac8{\pi a_0^3} \] Therefore, \[ \frac{\rho_{\mathrm{He^+}}(0)}{\rho_{\mathrm H}(0)}=8 \] \(\textbf{Construct the density ratio at an arbitrary radius:}\) At the same radial distance \(r\), \[ \frac{\rho_{\mathrm{He^+}}(r)}{\rho_{\mathrm H}(r)} =\frac{8e^{-4r/a_0}}{e^{-2r/a_0}} \] \[ \frac{\rho_{\mathrm{He^+}}(r)}{\rho_{\mathrm H}(r)} =8e^{-2r/a_0} \] \(\textbf{Find the radius at which the densities become equal:}\) Equality requires \[ 8e^{-2r/a_0}=1 \] Therefore, \[ e^{-2r/a_0}=\frac18 \] Taking natural logarithms, \[ -\frac{2r}{a_0}=-\ln8 \] \[ r=\frac{a_0\ln8}{2} \] Substituting, \[ r=\frac{2.079a_0}{2} \] \[ r\approx1.040a_0 \] \(\textbf{Compare ground-state binding energies:}\) For a hydrogen-like species, \[ |E_n|=13.6\frac{Z^2}{n^2}\,\mathrm{eV} \] Both systems are in \(n=1\). Therefore, \[ \frac{|E_{\mathrm{He^+}}|}{|E_{\mathrm H}|}=\frac{2^2}{1^2} \] \[ \frac{|E_{\mathrm{He^+}}|}{|E_{\mathrm H}|}=4 \] \(\textbf{Interpret the crossing of local densities:}\) The larger nuclear charge produces a more concentrated ground-state electron distribution. The \(\mathrm{He^+}\) density is eight times the hydrogen density at the nucleus. However, the stronger radial exponential decay allows the hydrogen density to become greater beyond the calculated crossing radius. The stronger binding of \(\mathrm{He^+}\) is consistent with its more compact spatial distribution. Local density equality at one radius does not imply equality of the complete wave functions or their energies. \(\textbf{Final answer:}\) \[ \boxed{\frac{\rho_{\mathrm{He^+}}(0)}{\rho_{\mathrm H}(0)}=8,\qquad r_{\mathrm{equal}}\approx1.040a_0,\qquad\frac{|E_{\mathrm{He^+}}|}{|E_{\mathrm H}|}=4} \]

Explanation: \(\textbf{Distinguish local finiteness from total normalisation:}\) The probability density associated with a wave function is its squared magnitude. For a normalised bound-state orbital, the integral of this density over all space must equal one. A function being finite at each individual point does not establish that its density has a finite total integral. The amount of spatial volume available at increasing radial distance must also be considered. \(\textbf{Examine the large-distance behaviour of P:}\) For P, \[ \Psi_P(r)=Ae^{-r/a} \] As \(r\) becomes large, \[ \Psi_P(r)\rightarrow0 \] Its probability density decreases as \[ |\Psi_P(r)|^2=A^2e^{-2r/a} \] The exponential decay is sufficiently rapid that the density can be integrated over all space to obtain a finite result. An appropriate choice of the overall normalisation constant can therefore make its total probability one. \(\textbf{Examine the large-distance behaviour of Q:}\) For Q, \[ \Psi_Q(r)=A(1+e^{-r/a}) \] At large radial distance, \[ \Psi_Q(r)\rightarrow A \] Therefore, \[ |\Psi_Q(r)|^2\rightarrow A^2 \] The density approaches a positive constant rather than zero. \(\textbf{Apply the spherical-volume requirement:}\) The volume of a thin spherical shell is approximately \(4\pi r^2dr\). For Q at sufficiently large radius, its probability contribution behaves like \[ dP\propto4\pi A^2r^2dr \] The available spherical-shell volume grows with radius. Accumulating these contributions to arbitrarily large distances produces a divergent total rather than a finite probability integral. Q cannot consequently be normalised over all space while \(A\neq0\). \(\textbf{Interpret the conclusion:}\) The proposed Q function is locally finite but fails the global probability requirement. P passes this necessary normalisation test. However, being normalisable is not, by itself, proof that a proposed function is an allowed stationary atomic orbital. An actual stationary orbital must also satisfy the appropriate quantum mechanical equation and physical boundary conditions. The investigator's error is to replace the total-probability requirement with the much weaker condition of finiteness at individual positions.

Explanation: The boundary surface is stipulated to enclose \(90\%\) of the total electron-position probability. It consists of two mirror-related lobes. For a real \(2p_z\) orbital, the spatial probability density is symmetric under reflection through the \(xy\)-plane. The probability associated with the part of the orbital on the positive-\(z\) side is consequently equal to that on the negative-\(z\) side. The two drawn lobes therefore divide the enclosed \(90\%\) probability equally. Each contains \(45\%\) of the total probability. The probability outside the complete drawn boundary is \(10\%\). The boundary is not an impenetrable wall; it is a chosen surface for representing a large fraction of the electron cloud. The real \(p_z\) wave function conventionally has opposite signs in its two principal lobes. However, the squared magnitude of the wave function gives the local probability density. Squaring removes the sign of a real wave-function amplitude. Both lobes consequently have nonnegative probability densities. The intervening \(xy\)-plane is an angular nodal plane on which the wave function vanishes. That zero-density plane is not a location where the unrepresented \(10\%\) probability is stored. The two lobes also do not constitute two independent spatial orbitals. Together they describe one \(2p_z\) spatial wave function. The experiment therefore distinguishes the wave-function sign, the symmetry of probability density and the probabilistic meaning of an orbital boundary surface.

Explanation: \(\textbf{Determine the wavelengths during passage through glass:}\) The refractive index is \[ n=\frac cv \] Therefore, the speed of light in glass is \(v=c/n\). Frequency remains unchanged at the boundary, while wavelength changes according to \[ \lambda_{\mathrm{glass}}=\frac{\lambda_{\mathrm{vac}}}{n} \] For P, \[ \lambda_{P,\mathrm{glass}}=\frac{450}{1.50} \] \[ \lambda_{P,\mathrm{glass}}=300\,\mathrm{nm} \] For Q, \[ \lambda_{Q,\mathrm{glass}}=\frac{350}{1.50} \] \[ \lambda_{Q,\mathrm{glass}}\approx233\,\mathrm{nm} \] \(\textbf{Identify the quantity governing the photoelectric threshold:}\) Photon energy is determined by its frequency: \[ E_\gamma=hf \] Refraction changes wavelength and propagation speed, but not the photon frequency. Therefore, passage through the transparent glass does not change the photon energy. The vacuum threshold wavelength must be compared with the beams' vacuum wavelengths or, equivalently, their unchanged frequencies. \(\textbf{Calculate the metal's work function:}\) \[ \phi=\frac{hc}{\lambda_0} \] \[ \phi=\frac{1240}{400} \] \[ \phi=3.10\,\mathrm{eV} \] \(\textbf{Test beam P:}\) \[ E_P=\frac{1240}{450} \] \[ E_P\approx2.756\,\mathrm{eV} \] Since \[ E_P\lt \phi \] beam P cannot cause photoemission. Its wavelength becoming \(300\,\mathrm{nm}\) inside glass does not raise its photon energy above threshold. \(\textbf{Test beam Q:}\) \[ E_Q=\frac{1240}{350} \] \[ E_Q\approx3.543\,\mathrm{eV} \] Since \(E_Q\gt \phi\), photoemission is possible. The maximum kinetic energy is \[ K_{\max}=E_Q-\phi \] \[ K_{\max}=3.543-3.100 \] \[ K_{\max}\approx0.443\,\mathrm{eV} \] \(\textbf{Interpret the changed propagation medium:}\) A photon does not become more energetic merely because its wavelength is shorter in a medium of lower propagation speed. The relation \(E=hc/\lambda\) uses the corresponding vacuum wavelength when \(c\) is the vacuum speed of light. Inside glass, the appropriate wave relation is \(v=f\lambda_{\mathrm{glass}}\). The photoelectric outcome depends on photon energy, not on comparing a wavelength measured inside glass directly with a threshold wavelength defined in vacuum. \(\textbf{Final answer:}\) \[ \boxed{\lambda_{P,\mathrm{glass}}=300\,\mathrm{nm},\quad\lambda_{Q,\mathrm{glass}}\approx233\,\mathrm{nm},\quad K_{\max,Q}\approx0.443\,\mathrm{eV}} \]

Explanation: \(\textbf{Evaluate the elementary Bohr angular momentum:}\) Bohr's quantisation rule assigns \[ L_{\mathrm{Bohr}}=n\hbar \] For the second orbit, \[ L_{\mathrm{Bohr},2}=2\hbar \] This angular momentum is associated with the circular-orbit model. \(\textbf{Evaluate the quantum mechanical }2s\textbf{ orbital:}\) An \(s\) orbital has azimuthal quantum number \[ l=0 \] The quantum mechanical orbital angular momentum magnitude is \[ L_{\mathrm{orb}}=\sqrt{l(l+1)}\,\hbar \] Therefore, \[ L_{2s}=\sqrt{0(0+1)}\,\hbar \] \[ L_{2s}=0 \] \(\textbf{Evaluate the quantum mechanical }2p\textbf{ orbital:}\) A \(p\) orbital has \[ l=1 \] Thus, \[ L_{2p}=\sqrt{1(1+1)}\,\hbar \] \[ L_{2p}=\sqrt2\,\hbar \] \(\textbf{Compare the electronic energies separately:}\) For hydrogen in the elementary one-electron treatment, \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] Both \(2s\) and \(2p\) have \(n=2\). Therefore, \[ E_{2s}=E_{2p}=-\frac{13.6}{4} \] \[ E_{2s}=E_{2p}=-3.40\,\mathrm{eV} \] \(\textbf{Interpret the difference between the models:}\) The principal quantum number in Bohr's circular-orbit rule is not interchangeable with the azimuthal quantum number in the quantum mechanical orbital-angular-momentum relation. The \(2s\) and \(2p\) spatial states can have different orbital angular momenta while remaining degenerate in the elementary hydrogen energy model. A zero value of orbital angular momentum for \(2s\) also does not mean that its electron lacks intrinsic spin. The comparison demonstrates why replacing an orbital probability distribution by a sharply defined Bohr circular path changes more than the drawing of the electron's location. \(\textbf{Final answer:}\) \[ \boxed{L_{\mathrm{Bohr},2}=2\hbar,\qquad L_{2s}=0,\qquad L_{2p}=\sqrt2\,\hbar} \]

Explanation: The ensemble is prepared as a statistical mixture of two kinds of hydrogen atoms. Half occupy \(1s\) and half occupy \(2s\). Its average local probability density is therefore the weighted sum of the separate orbital densities given in the stem. The \(2s\) orbital has one finite spherical radial node. At \(r=r_0\), its wave function vanishes. Consequently, \[ \rho_{2s}(r_0)=0 \] The \(1s\) orbital has no finite radial node. Its spatial probability density is positive at every finite radial distance. Therefore, \[ \rho_{1s}(r_0)\gt 0 \] \(\textbf{Evaluate the combined local density:}\) \[ \rho_{\mathrm{ens}}(r_0) =\frac12\rho_{1s}(r_0)+\frac12(0) \] \[ \rho_{\mathrm{ens}}(r_0)=\frac12\rho_{1s}(r_0)\gt 0 \] \(\textbf{Convert the ensemble density to radial probability:}\) For a spherically symmetric density, the radial probability distribution is \[ P(r)=4\pi r^2\rho(r) \] At the finite positive node radius, \[ P_{\mathrm{ens}}(r_0)=4\pi r_0^2\frac12\rho_{1s}(r_0) \] Since \(r_0\gt 0\) and \(\rho_{1s}(r_0)\gt 0\), \[ P_{\mathrm{ens}}(r_0)\gt 0 \] \(\textbf{Interpret the result:}\) A nodal surface belongs to a particular orbital wave function. It is not automatically a nodal surface of a density formed by combining contributions from different occupied or prepared states. The \(2s\) contribution vanishes at \(r_0\), but the \(1s\) contribution remains positive. The ensemble therefore does not display the complete radial-density zero of the pure \(2s\) state at that radius. This is an average of probabilities for separately prepared atoms, not an instruction to add their real wave-function amplitudes before squaring. Confusing those two operations would incorrectly introduce interference between the separately prepared components. The example distinguishes a pure orbital's nodal structure from the probability density observed in a mixed population. \(\textbf{Final answer:}\) \[ \rho_{\mathrm{ens}}(r_0)=\frac12\rho_{1s}(r_0)\gt 0,\qquad P_{\mathrm{ens}}(r_0)\gt 0 \]

Explanation: Bohr's radius relation for hydrogen is \[ r_n=a_0n^2 \] For \(n=3\), it gives \[ r_3=9a_0 \] This is the radius assigned to the third stationary circular orbit within Bohr's model. The quantum mechanical model instead describes the electron through a spatial orbital wave function. Its probability density is the squared magnitude of that wave function. The \(3s\), \(3p\) and \(3d\) orbitals have the same principal quantum number but different azimuthal quantum numbers. Their angular structures consequently differ. The \(3s\) orbital is spherically symmetric and has two finite radial nodes. A \(3p\) orbital has one angular node and one finite radial node. A \(3d\) orbital has two angular nodes and no finite radial node. Thus, their spatial probability distributions are not identical. Within the elementary hydrogen-like model, their energies are nevertheless equal because the energy depends on \(n\). Energy degeneracy does not require identical wave functions or orbital shapes. A customary \(90\%\) boundary surface encloses a selected fraction of electron-position probability. Its geometry must be related to the actual probability distribution. The circular Bohr radius does not itself specify the location or geometry of that boundary. In particular, the \(3p\) and \(3d\) orbital shapes are directional rather than spherically symmetric. Representing all three by identical spherical boundaries at \(9a_0\) therefore confuses two distinct atomic models. The legitimate conclusion is that equal \(n\) can imply equal gross electronic energy for hydrogen without implying one common, sharply defined electron trajectory or probability boundary.

Explanation: \(\textbf{Apply the nuclear-charge scaling of the density:}\) For hydrogen, \[ Z_{\mathrm H}=1 \] For the one-electron helium ion, \[ Z_{\mathrm{He^+}}=2 \] The supplied scaling law becomes \[ \rho_{\mathrm{He^+}}(r)=8\rho_{\mathrm H}(2r) \] \(\textbf{Compare densities at corresponding scaled radii:}\) At the helium-ion position \(r=R/2\), \[ \rho_{\mathrm{He^+}}(R/2)=8\rho_{\mathrm H}(R) \] Therefore, \[ \frac{\rho_{\mathrm{He^+}}(R/2)}{\rho_{\mathrm H}(R)}=8 \] \(\textbf{Determine the probability enclosed by a scaled sphere:}\) For a spherical density, the enclosed probability is \[ F_Z(r)=4\pi\int_0^r r'^2\rho_Z(r')\,dr' \] Using the supplied scaling relation, the substitution \(u=Zr'\) gives \[ F_Z(r)=F_1(Zr) \] Thus, corresponding radii related by the factor \(1/Z\) enclose equal probabilities. Because \[ F_1(R)=0.90 \] the helium-ion sphere satisfying the same condition obeys \[ F_2(R/2)=F_1(R) \] \[ F_2(R/2)=0.90 \] Therefore, its \(90\%\)-probability radius is \(R/2\). \(\textbf{Compare the corresponding thin-shell volumes:}\) For a sufficiently thin spherical shell, \[ \Delta V\approx4\pi r^2\Delta r \] The hydrogen shell has radius \(R\) and thickness \(\Delta r\). The corresponding helium-ion shell has radius \(R/2\) and thickness \(\Delta r/2\). Their volume ratio is \[ \frac{\Delta V_{\mathrm{He^+}}}{\Delta V_{\mathrm H}} =\frac{(R/2)^2(\Delta r/2)}{R^2\Delta r} \] \[ \frac{\Delta V_{\mathrm{He^+}}}{\Delta V_{\mathrm H}}=\frac18 \] \(\textbf{Combine volume and density changes:}\) The corresponding helium-ion density is eight times greater. Its shell volume is one-eighth as large. Therefore, \[ \frac{\Delta P_{\mathrm{He^+}}}{\Delta P_{\mathrm H}} =8\left(\frac18\right) \] \[ \frac{\Delta P_{\mathrm{He^+}}}{\Delta P_{\mathrm H}}=1 \] \(\textbf{Interpret the probability scaling:}\) The stronger nuclear attraction contracts the hydrogen-like \(1s\) probability distribution. The local density rises as the occupied spatial region contracts. However, the corresponding scaled spherical regions contain equal integrated probabilities. Thus, a smaller boundary radius and a greater local density are consistent with normalisation. The result uses the complete spatial scaling of the density rather than equating a Bohr orbit radius directly with an orbital boundary. \(\textbf{Final answer:}\) \[ \boxed{R_{90,\mathrm{He^+}}=\frac R2,\qquad\frac{\rho_{\mathrm{He^+}}(R/2)}{\rho_{\mathrm H}(R)}=8,\qquad\frac{\Delta P_{\mathrm{He^+}}}{\Delta P_{\mathrm H}}=1} \]

Explanation: \(\textbf{Infer the electron's initial kinetic energy:}\) The initial electron de Broglie wavelength is \[ \lambda_{e,i}=0.200\,\mathrm{nm} \] Using the supplied relation, \[ K_{e,i}=\left(\frac{1.227}{\lambda_{e,i}(\mathrm{nm})}\right)^2\mathrm{eV} \] Substituting, \[ K_{e,i}=\left(\frac{1.227}{0.200}\right)^2 \] \[ K_{e,i}\approx37.64\,\mathrm{eV} \] \(\textbf{Apply the changed electron kinetic energy:}\) The final energy is four times the initial value: \[ K_{e,f}=4K_{e,i} \] \[ K_{e,f}\approx4(37.64) \] \[ K_{e,f}\approx150.55\,\mathrm{eV} \] The kinetic-energy gain is \[ \Delta K_e=K_{e,f}-K_{e,i} \] \[ \Delta K_e\approx150.55-37.64 \] \[ \Delta K_e\approx112.91\,\mathrm{eV} \] \(\textbf{Determine the accelerating potential difference:}\) For an electron accelerated through an appropriate potential difference of magnitude \(V\), \[ \Delta K_e=eV \] Expressed in electronvolts, \[ V\approx112.9\,\mathrm{V} \] \(\textbf{Determine the electron's new wavelength:}\) For a non-relativistic electron, \[ \lambda_e\propto\frac1{\sqrt{K_e}} \] Therefore, \[ \frac{\lambda_{e,f}}{\lambda_{e,i}}=\sqrt{\frac{K_{e,i}}{K_{e,f}}} \] \[ \frac{\lambda_{e,f}}{\lambda_{e,i}}=\frac12 \] \[ \lambda_{e,f}=0.100\,\mathrm{nm} \] \(\textbf{Evaluate the neutron's response independently:}\) The neutron possesses zero net electric charge. Under the stated ideal conditions, the electrostatic field does not accelerate it. Its central momentum magnitude therefore remains unchanged. Consequently, \[ \lambda_{n,f}=0.200\,\mathrm{nm} \] \(\textbf{Compare the final matter wavelengths:}\) \[ \frac{\lambda_{n,f}}{\lambda_{e,f}}=\frac{0.200}{0.100} \] \[ \frac{\lambda_{n,f}}{\lambda_{e,f}}=2 \] \(\textbf{Interpret the different physical principles:}\) Both electrons and neutrons exhibit matter-wave behaviour because de Broglie's relation applies to moving material particles. However, their response to an electrostatic accelerating field depends on electric charge. The neutron does not lose its matter-wave character merely because its wavelength remains unchanged. The electron's shorter final wavelength results from increased momentum, whereas the neutron retains its original momentum under the stipulated conditions. This separates the general existence of matter waves from the specific force used to change a particle's motion. \(\textbf{Final answer:}\) \[ \boxed{V\approx112.9\,\mathrm{V},\qquad\lambda_{e,f}=0.100\,\mathrm{nm},\qquad\frac{\lambda_{n,f}}{\lambda_{e,f}}=2} \]