Dual Nature of Matter and Radiation MCQs with Answers – Part 4 (NEET Physics)

Explanation: \(\textbf{Infer the momentum ratio from wavelengths:}\) De Broglie's relation gives: \[ p=\frac{h}{\lambda} \] For P: \[ p_P=\frac{h}{\lambda} \] For Q: \[ p_Q=\frac{h}{\lambda/2} \] \[ p_Q=2p_P \] Thus: \[ \frac{p_Q}{p_P}=2 \] \(\textbf{Relate momentum and kinetic energy to mass:}\) For non-relativistic motion: \[ K=\frac{p^2}{2m} \] Rearranging: \[ m=\frac{p^2}{2K} \] Therefore: \[ \frac{m_Q}{m_P}=\left(\frac{p_Q}{p_P}\right)^2\frac{K_P}{K_Q} \] Substituting: \[ \frac{m_Q}{m_P}=2^2\left(\frac{1}{4}\right) \] \[ \frac{m_Q}{m_P}=1 \] Hence: \[ m_Q=m_P \] \(\textbf{Recover the speed ratio:}\) Since: \[ p=mv \] we have: \[ \frac{v_Q}{v_P}=\frac{p_Q}{p_P}\frac{m_P}{m_Q} \] Substituting: \[ \frac{v_Q}{v_P}=2(1) \] \[ \frac{v_Q}{v_P}=2 \] \(\textbf{Physical interpretation:}\) The measured wavelengths independently reveal the particle momentum ratio. The kinetic-energy measurements then determine whether that momentum ratio can be explained by equal or unequal masses. Here, the fourfold energy ratio and twofold momentum ratio together imply equal masses. The speed of Q is consequently twice that of P. \(\textbf{Final answer:}\) \[ \boxed{\frac{m_Q}{m_P}=1,\qquad\frac{v_Q}{v_P}=2} \]

Explanation: \(\textbf{Identify the physical graph slope:}\) The de Broglie relation is: \[ \lambda=\frac{h}{p} \] Equivalently: \[ \lambda=h\left(\frac{1}{p}\right) \] Therefore, the slope of a wavelength-versus-reciprocal-momentum graph is Planck's constant when SI units are used. \(\textbf{Convert the displayed graph slope into SI units:}\) The plotted numerical slope is: \[ s=0.663\ \frac{\text{nm}}{10^{24}\,\text{s kg}^{-1}\text{m}^{-1}} \] Convert nanometres into metres: \[ 0.663\,\text{nm}=0.663\times10^{-9}\,\text{m} \] Thus: \[ h=\frac{0.663\times10^{-9}\,\text{m}}{10^{24}\,\text{s kg}^{-1}\text{m}^{-1}} \] \[ h=0.663\times10^{-33}\,\text{kg m}^2\text{s}^{-1} \] \[ h=6.63\times10^{-34}\,\text{J s} \] \(\textbf{Determine momentum from the measured wavelength:}\) The particle's wavelength is: \[ \lambda=0.221\,\text{nm} \] Convert into SI units: \[ \lambda=0.221\times10^{-9}\,\text{m} \] Using: \[ p=\frac{h}{\lambda} \] we obtain: \[ p=\frac{6.63\times10^{-34}}{0.221\times10^{-9}} \] \[ p=3.00\times10^{-24}\,\text{kg m s}^{-1} \] \(\textbf{Dimensional and graphical check:}\) The ratio of wavelength to reciprocal momentum has the dimensions of momentum multiplied by length. These are the dimensions of action, matching Planck's constant. The particle with wavelength \(0.221\,\text{nm}\) must have greater momentum than one with wavelength \(0.663\,\text{nm}\), consistent with the inverse relationship. \(\textbf{Final answer:}\) \[ \boxed{h=6.63\times10^{-34}\,\text{J s},\qquad p=3.00\times10^{-24}\,\text{kg m s}^{-1}} \]

Explanation: \(\textbf{Write the initial momentum vector:}\) The particle initially moves along the positive \(x\)-axis. Therefore: \[ \vec p_i=p\hat i \] The associated wavelength is: \[ \lambda_0=\frac{h}{p} \] \(\textbf{Apply the perpendicular impulse:}\) Impulse equals the change in momentum: \[ \vec J=\Delta\vec p \] The applied impulse is: \[ \vec J=2p\hat j \] Therefore: \[ \vec p_f=\vec p_i+\vec J \] Substituting: \[ \vec p_f=p\hat i+2p\hat j \] \[ \vec p_f=p(\hat i+2\hat j) \] \(\textbf{Calculate final momentum magnitude:}\) The two momentum components are perpendicular. Thus: \[ p_f=\sqrt{p_x^2+p_y^2} \] \[ p_f=\sqrt{p^2+(2p)^2} \] \[ p_f=\sqrt{5}\,p \] \(\textbf{Determine the new matter wavelength:}\) Using: \[ \lambda_f=\frac{h}{p_f} \] we obtain: \[ \lambda_f=\frac{h}{\sqrt{5}\,p} \] Since: \[ \lambda_0=\frac{h}{p} \] therefore: \[ \lambda_f=\frac{\lambda_0}{\sqrt{5}} \] \(\textbf{Determine the final momentum direction:}\) Let \(\theta\) be the angle measured from the positive \(x\)-axis towards the positive \(y\)-axis. Then: \[ \tan\theta=\frac{p_y}{p_x} \] \[ \tan\theta=\frac{2p}{p} \] \[ \tan\theta=2 \] Therefore: \[ \theta=\tan^{-1}(2) \] \(\textbf{Physical interpretation:}\) The de Broglie wavelength depends on the magnitude of the resultant momentum vector. Because the impulse is perpendicular to the initial momentum, the magnitudes cannot simply be added arithmetically. The impulse changes both the direction and magnitude of momentum, and hence changes the matter wavelength. \(\textbf{Final answer:}\) \[ \boxed{\lambda_f=\frac{\lambda_0}{\sqrt{5}},\qquad\theta=\tan^{-1}(2)} \]

Explanation: \(\textbf{Relate accelerating voltage to matter wavelength:}\) For a particle accelerated from rest through a suitable potential difference: \[ K=|q|V \] For non-relativistic motion: \[ p=\sqrt{2mK} \] Therefore: \[ p=\sqrt{2m|q|V} \] Using the de Broglie relation: \[ \lambda=\frac{h}{\sqrt{2m|q|V}} \] Rearranging: \[ \lambda=\frac{h}{\sqrt{2m|q|}}\left(\frac{1}{\sqrt V}\right) \] \(\textbf{Interpret the graph slopes:}\) The slope is: \[ s=\frac{h}{\sqrt{2m|q|}} \] Consequently: \[ \frac{s_P}{s_Q}=\sqrt{\frac{m_Q|q_Q|}{m_P|q_P|}} \] The measured slope ratio is: \[ \frac{s_P}{s_Q}=3 \] Squaring: \[ 9=\frac{m_Q|q_Q|}{m_P|q_P|} \] \(\textbf{Use the independently measured mass ratio:}\) The problem gives: \[ m_Q=3m_P \] Therefore: \[ 9=\frac{3m_P|q_Q|}{m_P|q_P|} \] \[ 9=3\frac{|q_Q|}{|q_P|} \] Hence: \[ \frac{|q_Q|}{|q_P|}=3 \] \(\textbf{Compare speeds at equal accelerating voltage:}\) Using: \[ \frac{1}{2}mv^2=|q|V \] we obtain: \[ v=\sqrt{\frac{2|q|V}{m}} \] At equal \(V\): \[ \frac{v_Q}{v_P}=\sqrt{\frac{|q_Q|m_P}{|q_P|m_Q}} \] Substituting the inferred charge and given mass ratios: \[ \frac{v_Q}{v_P}=\sqrt{\frac{3}{3}} \] \[ \frac{v_Q}{v_P}=1 \] \(\textbf{Physical interpretation:}\) Although Q is three times heavier, it also carries three times the charge magnitude. The greater electrical energy supplied to Q exactly compensates for its greater mass in the speed comparison. Its momentum is nevertheless greater, so its de Broglie wavelength is smaller. The graph slopes contain information about the combined mass-charge product, not mass or charge alone. \(\textbf{Final answer:}\) \[ \boxed{\frac{|q_Q|}{|q_P|}=3,\qquad\frac{v_Q}{v_P}=1} \]

Explanation: \(\textbf{Energy supplied during photoelectric emission:}\) The incident photon energy is: \[ E_\gamma=4.0\,\text{eV} \] The metal work function is: \[ \phi=3.0\,\text{eV} \] Einstein's photoelectric equation gives: \[ K_i=E_\gamma-\phi \] Substituting: \[ K_i=4.0-3.0 \] \[ K_i=1.0\,\text{eV} \] \(\textbf{Determine the independent stopping potential:}\) The stopping-potential magnitude satisfies: \[ eV_0=K_i \] Therefore: \[ V_0=1.0\,\text{V} \] This value describes the energy immediately after emission. \(\textbf{Include subsequent acceleration:}\) The collector is maintained at: \[ V_C=+8.0\,\text{V} \] The electron gains: \[ \Delta K=eV_C \] \[ \Delta K=8.0\,\text{eV} \] Thus: \[ K_f=K_i+\Delta K \] \[ K_f=1.0+8.0 \] \[ K_f=9.0\,\text{eV} \] \(\textbf{Compare electron matter wavelengths:}\) For the same non-relativistic electron: \[ \lambda=\frac{h}{\sqrt{2m_eK}} \] Therefore: \[ \frac{\lambda_f}{\lambda_i}=\sqrt{\frac{K_i}{K_f}} \] Substituting: \[ \frac{\lambda_f}{\lambda_i}=\sqrt{\frac{1.0}{9.0}} \] \[ \frac{\lambda_f}{\lambda_i}=\frac{1}{3} \] \(\textbf{Physical interpretation:}\) The positive collector potential supplies additional kinetic energy after the electron has escaped from the metal. This increases electron momentum and decreases its associated de Broglie wavelength. However, the stopping potential obtained in a separate photoelectric measurement still depends on the original maximum kinetic energy immediately after emission. The accelerating and emission stages must not be confused. \(\textbf{Final answer:}\) \[ \boxed{\frac{\lambda_f}{\lambda_i}=\frac{1}{3},\qquad V_0=1.0\,\text{V}} \]

Explanation: \(\textbf{Determine electrical work for each charge:}\) The potential difference is: \[ \Delta V=+5.0\,\text{V} \] The change in kinetic energy of a particle of charge \(q\) is: \[ \Delta K=-q\Delta V \] For an electron: \[ q_e=-e \] Therefore: \[ \Delta K_e=+e(5.0\,\text{V}) \] \[ \Delta K_e=+5.0\,\text{eV} \] For a proton: \[ q_p=+e \] Therefore: \[ \Delta K_p=-e(5.0\,\text{V}) \] \[ \Delta K_p=-5.0\,\text{eV} \] \(\textbf{Calculate final kinetic energies:}\) Both initially possess: \[ K_i=8.0\,\text{eV} \] Hence: \[ K_{e,f}=8.0+5.0 \] \[ K_{e,f}=13.0\,\text{eV} \] For the proton: \[ K_{p,f}=8.0-5.0 \] \[ K_{p,f}=3.0\,\text{eV} \] The proton still has positive kinetic energy and can reach the second electrode. \(\textbf{Compare matter wavelengths:}\) The non-relativistic de Broglie relation is: \[ \lambda=\frac{h}{\sqrt{2mK}} \] Therefore: \[ \frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_pK_{p,f}}{m_eK_{e,f}}} \] Substituting: \[ \frac{\lambda_e}{\lambda_p}=\sqrt{\frac{1836m_e(3.0)}{m_e(13.0)}} \] \[ \frac{\lambda_e}{\lambda_p}=\sqrt{\frac{5508}{13}} \] \[ \frac{\lambda_e}{\lambda_p}\approx20.6 \] \(\textbf{Physical interpretation:}\) Moving towards higher electric potential accelerates the negative electron but retards the positive proton. Their final kinetic energies therefore change in opposite directions. Their de Broglie wavelengths must be compared using both their final energies and their different masses. Equal initial kinetic energies do not guarantee equal wavelengths either before or after the electrical interaction. \(\textbf{Final answer:}\) \[ \boxed{\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{5508}{13}}\approx20.6} \]

Explanation: \(\textbf{Compare kinetic-energy gains:}\) A charged particle accelerated from rest gains: \[ K=|q|V \] For the proton: \[ K_p=eV \] For the alpha particle: \[ K_\alpha=2eV \] Therefore: \[ \frac{K_\alpha}{K_p}=2 \] \(\textbf{Determine momentum ratio:}\) For non-relativistic motion: \[ p=\sqrt{2mK} \] Thus: \[ \frac{p_\alpha}{p_p}=\sqrt{\frac{m_\alpha K_\alpha}{m_pK_p}} \] Using: \[ m_\alpha=4m_p \] and: \[ K_\alpha=2K_p \] we obtain: \[ \frac{p_\alpha}{p_p}=\sqrt{4(2)} \] \[ \frac{p_\alpha}{p_p}=2\sqrt{2} \] \(\textbf{Compare associated wavelengths:}\) Since: \[ \lambda=\frac{h}{p} \] we have: \[ \frac{\lambda_\alpha}{\lambda_p}=\frac{p_p}{p_\alpha} \] Therefore: \[ \frac{\lambda_\alpha}{\lambda_p}=\frac{1}{2\sqrt{2}} \] \(\textbf{Compare final speeds:}\) From: \[ K=\frac{1}{2}mv^2 \] the speed is: \[ v=\sqrt{\frac{2K}{m}} \] Therefore: \[ \frac{v_\alpha}{v_p}=\sqrt{\frac{K_\alpha m_p}{K_pm_\alpha}} \] Substituting: \[ \frac{v_\alpha}{v_p}=\sqrt{\frac{2}{4}} \] \[ \frac{v_\alpha}{v_p}=\frac{1}{\sqrt{2}} \] \(\textbf{Physical interpretation:}\) The alpha particle gains twice as much kinetic energy because its charge magnitude is twice as large. Nevertheless, its mass is four times the proton mass, so its final speed is smaller. Its momentum is larger, however, and consequently its de Broglie wavelength is shorter. Speed and momentum comparisons therefore lead to different orderings for these particles. \(\textbf{Final answer:}\) \[ \boxed{\frac{\lambda_\alpha}{\lambda_p}=\frac{1}{2\sqrt{2}},\qquad\frac{v_\alpha}{v_p}=\frac{1}{\sqrt{2}}} \]

Explanation: \(\textbf{Identify the undeflected-particle condition:}\) An electron moving through crossed electric and magnetic fields experiences electric and magnetic forces. For a straight trajectory, the forces must be equal in magnitude and opposite in direction. Therefore: \[ |q|E=|q|vB \] Cancelling charge magnitude: \[ E=vB \] Thus: \[ v=\frac{E}{B} \] \(\textbf{Calculate the selected electron speed:}\) Substituting: \[ v=\frac{3.0\times10^4}{5.0\times10^{-2}}\,\text{m s}^{-1} \] \[ v=6.0\times10^5\,\text{m s}^{-1} \] \(\textbf{Determine the associated matter wavelength:}\) For a non-relativistic electron: \[ p=m_ev \] Therefore: \[ \lambda=\frac{h}{m_ev} \] Substituting: \[ \lambda=\frac{6.63\times10^{-34}}{(9.10\times10^{-31})(6.0\times10^5)}\,\text{m} \] \[ \lambda=1.214\times10^{-9}\,\text{m} \] \[ \lambda\approx1.21\,\text{nm} \] \(\textbf{Apply the simultaneous field change:}\) The new field magnitudes are: \[ E'=2E \] \[ B'=2B \] The selected speed becomes: \[ v'=\frac{E'}{B'} \] \[ v'=\frac{2E}{2B} \] \[ v'=v \] Hence: \[ \lambda'=\frac{h}{m_ev'}=\lambda \] \(\textbf{Physical interpretation:}\) A crossed-field arrangement selects electron speed through the ratio of electric-field magnitude to magnetic-field magnitude. Doubling both fields increases their individual forces but leaves their balance at the same electron speed. The selected particles therefore retain the same momentum magnitude and de Broglie wavelength. A change in field strength alone does not guarantee a wavelength change. \(\textbf{Final answer:}\) \[ \boxed{\lambda\approx1.21\,\text{nm},\qquad\frac{\lambda'}{\lambda}=1} \]

Explanation: \(\textbf{Derive the graph equation:}\) The de Broglie relation gives: \[ \lambda=\frac{h}{p} \] For non-relativistic motion: \[ p^2=2mK \] Therefore: \[ \lambda^2=\frac{h^2}{2mK} \] Taking the reciprocal: \[ \frac{1}{\lambda^2}=\frac{2m}{h^2}K \] \(\textbf{Interpret the slope ratio:}\) The slope of the graph is: \[ s=\frac{2m}{h^2} \] Thus: \[ \frac{s_P}{s_Q}=\frac{m_P}{m_Q} \] The observed slope ratio is 9. Therefore: \[ \frac{m_P}{m_Q}=9 \] \(\textbf{Compare wavelengths at equal kinetic energy:}\) For fixed \(K\): \[ \lambda=\frac{h}{\sqrt{2mK}} \] Thus: \[ \lambda\propto\frac{1}{\sqrt m} \] Consequently: \[ \frac{\lambda_P}{\lambda_Q}=\sqrt{\frac{m_Q}{m_P}} \] \[ \frac{\lambda_P}{\lambda_Q}=\sqrt{\frac{1}{9}} \] \[ \frac{\lambda_P}{\lambda_Q}=\frac{1}{3} \] \(\textbf{Compare kinetic energies at equal wavelength:}\) Equal wavelength implies equal momentum magnitude. Using: \[ K=\frac{p^2}{2m} \] we obtain: \[ \frac{K_P}{K_Q}=\frac{m_Q}{m_P} \] Therefore: \[ \frac{K_P}{K_Q}=\frac{1}{9} \] \(\textbf{Physical interpretation:}\) The graph slope is directly proportional to mass, whereas wavelength at equal kinetic energy is inversely proportional to the square root of mass. At equal wavelength, the momenta are equal and the kinetic energies are inversely proportional to mass. These are three distinct comparisons, and applying one proportionality indiscriminately would produce incorrect results. \(\textbf{Final answer:}\) \[ \boxed{\frac{m_P}{m_Q}=9,\qquad\left(\frac{\lambda_P}{\lambda_Q}\right)_{K_P=K_Q}=\frac{1}{3},\qquad\left(\frac{K_P}{K_Q}\right)_{\lambda_P=\lambda_Q}=\frac{1}{9}} \]

Explanation: \(\textbf{Infer particle momenta from the common wavelength:}\) De Broglie's relation is: \[ \lambda=\frac{h}{p} \] The two wavelengths are equal. Therefore: \[ p_P=p_Q=p \] \(\textbf{Compare their speeds:}\) For non-relativistic motion: \[ p=mv \] Hence: \[ v=\frac{p}{m} \] The masses are: \[ m_P=5m \] \[ m_Q=m \] Therefore: \[ v_P=\frac{p}{5m} \] \[ v_Q=\frac{p}{m} \] Thus: \[ \frac{v_P}{v_Q}=\frac{1}{5} \] \(\textbf{Determine the flight-time ratio:}\) The flight distance is the same for both beams. Using: \[ t=\frac{L}{v} \] we obtain: \[ \frac{t_P}{t_Q}=\frac{v_Q}{v_P} \] \[ \frac{t_P}{t_Q}=5 \] \(\textbf{Compare kinetic energies:}\) At fixed momentum: \[ K=\frac{p^2}{2m} \] Therefore: \[ \frac{K_P}{K_Q}=\frac{m_Q}{m_P} \] \[ \frac{K_P}{K_Q}=\frac{1}{5} \] \(\textbf{Physical interpretation:}\) Equal de Broglie wavelengths indicate equal momentum magnitudes, not equal speeds. The heavier particle must move more slowly to possess the same momentum. Consequently, it requires more time to cross the same distance, while its kinetic energy is smaller. The time-of-flight measurements therefore provide an independent way to distinguish equal-momentum particles of different masses. \(\textbf{Final answer:}\) \[ \boxed{\frac{t_P}{t_Q}=5,\qquad\frac{K_P}{K_Q}=\frac{1}{5}} \]

Explanation: \(\textbf{Apply momentum conservation:}\) Initially, P carries momentum: \[ p_i=mv \] Particle Q is at rest, so its initial momentum is zero. After the perfectly inelastic collision, the composite object has mass: \[ M=2m \] Let its speed be \(u\). Conservation of momentum gives: \[ mv=(2m)u \] Therefore: \[ u=\frac{v}{2} \] \(\textbf{Compare initial and final momenta:}\) The composite momentum is: \[ p_f=(2m)\left(\frac{v}{2}\right) \] \[ p_f=mv \] Therefore: \[ p_f=p_i \] \(\textbf{Compare de Broglie wavelengths:}\) Using: \[ \lambda=\frac{h}{p} \] we obtain: \[ \frac{\lambda_f}{\lambda_i}=\frac{p_i}{p_f} \] \[ \frac{\lambda_f}{\lambda_i}=1 \] \(\textbf{Compare kinetic energies:}\) Initially: \[ K_i=\frac{1}{2}mv^2 \] Finally: \[ K_f=\frac{1}{2}(2m)u^2 \] Substituting \(u=v/2\): \[ K_f=\frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 \] \[ K_f=\frac{mv^2}{4} \] Therefore: \[ \frac{K_f}{K_i}=\frac{mv^2/4}{mv^2/2} \] \[ \frac{K_f}{K_i}=\frac{1}{2} \] \(\textbf{Physical interpretation:}\) Momentum is conserved in the collision even though kinetic energy is not. The lost translational kinetic energy is converted into other forms of energy during the sticking process. Because the composite object's momentum magnitude equals P's original momentum magnitude, its de Broglie wavelength is unchanged. Equal matter wavelengths therefore do not imply equal masses or equal kinetic energies. \(\textbf{Final answer:}\) \[ \boxed{\frac{\lambda_f}{\lambda_i}=1,\qquad\frac{K_f}{K_i}=\frac{1}{2}} \]

Explanation: \(\textbf{Determine the photoelectric emission energies:}\) For the unchanged photon energy: \[ E_\gamma=4.0\,\text{eV} \] The maximum kinetic energy after emission from P is: \[ K_{P,i}=E_\gamma-\phi_P \] \[ K_{P,i}=4.0-2.0 \] \[ K_{P,i}=2.0\,\text{eV} \] For Q: \[ K_{Q,i}=4.0-3.0 \] \[ K_{Q,i}=1.0\,\text{eV} \] \(\textbf{Compare photoelectric stopping potentials:}\) Since: \[ eV_0=K_{\max} \] we have: \[ V_{0P}=2.0\,\text{V} \] \[ V_{0Q}=1.0\,\text{V} \] Thus, the intrinsic stopping-potential magnitudes differ. \(\textbf{Include the later electrical acceleration:}\) For P, the additional kinetic energy is: \[ \Delta K_P=1.0\,\text{eV} \] Hence: \[ K_{P,f}=2.0+1.0 \] \[ K_{P,f}=3.0\,\text{eV} \] For Q: \[ \Delta K_Q=2.0\,\text{eV} \] Therefore: \[ K_{Q,f}=1.0+2.0 \] \[ K_{Q,f}=3.0\,\text{eV} \] \(\textbf{Compare final matter wavelengths:}\) For electrons of equal mass: \[ \lambda=\frac{h}{\sqrt{2m_eK}} \] Since: \[ K_{P,f}=K_{Q,f} \] their final momentum magnitudes are equal. Therefore: \[ \lambda_{P,f}=\lambda_{Q,f} \] \(\textbf{Physical interpretation:}\) The two photocells initially produce photoelectrons with different maximum kinetic energies because their work functions differ. Their subsequent accelerating potentials are chosen so that the final kinetic energies become equal. Consequently, their final de Broglie wavelengths are identical, even though their photoelectric stopping potentials remain different. This separates energy gained during emission from energy supplied later by an electric field. \(\textbf{Final answer:}\) \[ \boxed{K_{P,f}=K_{Q,f}=3.0\,\text{eV},\qquad\lambda_{P,f}=\lambda_{Q,f}} \]

Explanation: \(\textbf{Relate magnetic curvature to momentum:}\) A charged particle moving perpendicular to a uniform magnetic field experiences a magnetic force: \[ F_B=|q|vB \] For circular motion: \[ \frac{mv^2}{r}=|q|vB \] Cancelling one factor of speed: \[ mv=|q|Br \] Thus: \[ p=|q|Br \] \(\textbf{Compare momenta at equal field and radius:}\) For an electron: \[ |q_e|=e \] Therefore: \[ p_e=eBr \] For an alpha particle: \[ |q_\alpha|=2e \] Hence: \[ p_\alpha=2eBr \] Therefore: \[ \frac{p_\alpha}{p_e}=2 \] \(\textbf{Compare de Broglie wavelengths:}\) Using: \[ \lambda=\frac{h}{p} \] we obtain: \[ \frac{\lambda_\alpha}{\lambda_e}=\frac{p_e}{p_\alpha} \] \[ \frac{\lambda_\alpha}{\lambda_e}=\frac{1}{2} \] \(\textbf{Physical interpretation:}\) The circular-path radius depends on the ratio of momentum to charge magnitude. Equal radii in the same magnetic field therefore do not imply equal momenta when the particles carry different charge magnitudes. The alpha particle must have twice the momentum magnitude to follow the same radius. Its associated matter wavelength is consequently half that of the electron. Particle mass affects the speed and kinetic energy required to produce the specified momentum, but the wavelength comparison follows directly from momentum. \(\textbf{Final answer:}\) \[ \boxed{\frac{p_\alpha}{p_e}=2,\qquad\frac{\lambda_\alpha}{\lambda_e}=\frac{1}{2}} \]

Explanation: The assertion is correct because a perfectly elastic rebound from a stationary wall preserves the particle's kinetic energy under the stated idealisation. Since the particle mass is unchanged, its speed magnitude remains unchanged. However, the direction of motion reverses, so its momentum vector changes from its initial direction to the opposite direction. The magnitude of the momentum remains the same. De Broglie's relation associates wavelength with the magnitude of the particle's momentum, not with the sign of a chosen momentum component. Consequently, the particle's de Broglie wavelength magnitude is identical before and after the collision. The reason correctly identifies both the directional momentum change and the preservation of momentum magnitude. This example also demonstrates why a large change in momentum vector does not necessarily imply a change in matter wavelength. Momentum is a vector, whereas wavelength is a scalar length. The reflection changes the direction of propagation of the associated wave without requiring a different wavelength magnitude. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Explanation: \(\textbf{Relate reciprocal wavelength to momentum:}\) The de Broglie relation gives: \[ \lambda=\frac{h}{p} \] Therefore: \[ \frac{1}{\lambda}=\frac{p}{h} \] \(\textbf{Apply constant-force dynamics:}\) The particle moves along the force direction. For constant net force \(F\): \[ p=p_0+Ft \] Substituting into the reciprocal-wavelength relation: \[ \frac{1}{\lambda}=\frac{p_0}{h}+\frac{F}{h}t \] This is a straight-line equation. \(\textbf{Interpret the measured slope:}\) The slope is: \[ s=\frac{F}{h} \] Therefore: \[ F=hs \] Substituting: \[ F=(6.63\times10^{-34})(3.0\times10^{20})\,\text{N} \] \[ F=1.989\times10^{-13}\,\text{N} \] \[ F\approx1.99\times10^{-13}\,\text{N} \] \(\textbf{Use the initial intercept:}\) At: \[ t=0 \] the graph gives: \[ \left(\frac{1}{\lambda}\right)_0=5.0\times10^9\,\text{m}^{-1} \] Thus: \[ \lambda_0=\frac{1}{5.0\times10^9}\,\text{m} \] \[ \lambda_0=0.200\,\text{nm} \] \(\textbf{Calculate reciprocal wavelength after 10 ps:}\) The elapsed time is: \[ t=10\,\text{ps}=1.0\times10^{-11}\,\text{s} \] Using the line equation: \[ \frac{1}{\lambda}=\left(\frac{1}{\lambda}\right)_0+st \] Substituting: \[ \frac{1}{\lambda}=5.0\times10^9+(3.0\times10^{20})(1.0\times10^{-11}) \] \[ \frac{1}{\lambda}=5.0\times10^9+3.0\times10^9 \] \[ \frac{1}{\lambda}=8.0\times10^9\,\text{m}^{-1} \] Therefore: \[ \lambda=\frac{1}{8.0\times10^9}\,\text{m} \] \[ \lambda\approx1.25\times10^{-10}\,\text{m} \] \[ \lambda\approx0.125\,\text{nm} \] \(\textbf{Physical interpretation:}\) A constant force produces a linear change of momentum with time. Since reciprocal de Broglie wavelength is proportional to momentum, its graph against time is also linear while the motion remains along the force direction. The wavelength itself decreases nonlinearly as momentum increases. \(\textbf{Final answer:}\) \[ \boxed{F\approx1.99\times10^{-13}\,\text{N},\qquad\lambda(10\,\text{ps})\approx0.125\,\text{nm}} \]

Explanation: \(\textbf{Express kinetic-energy power:}\) Let the electron arrival rates be: \[ R_P=R \] \[ R_Q=4R \] Let their individual electron kinetic energies be \(K_P\) and \(K_Q\). The kinetic-energy power of a monoenergetic particle beam is: \[ P_K=RK \] \(\textbf{Apply the equal-power condition:}\) The two beams have equal kinetic-energy powers: \[ R_PK_P=R_QK_Q \] Substituting: \[ RK_P=(4R)K_Q \] Cancelling \(R\): \[ K_P=4K_Q \] Therefore: \[ \frac{K_Q}{K_P}=\frac{1}{4} \] \(\textbf{Compare matter wavelengths:}\) For non-relativistic electrons: \[ \lambda=\frac{h}{\sqrt{2m_eK}} \] Since the particle masses are identical: \[ \frac{\lambda_Q}{\lambda_P}=\sqrt{\frac{K_P}{K_Q}} \] Substituting: \[ \frac{\lambda_Q}{\lambda_P}=\sqrt{4} \] \[ \frac{\lambda_Q}{\lambda_P}=2 \] \(\textbf{Physical interpretation:}\) Equal total kinetic-energy power does not require equal kinetic energy per electron. Beam Q contains four times as many electrons per second, so each Q electron carries one-quarter of the kinetic energy of a P electron. Its momentum is consequently half as great, and its associated de Broglie wavelength is twice as long. Beam intensity and individual particle wavelength are therefore distinct quantities. \(\textbf{Final answer:}\) \[ \boxed{\frac{R_Q}{R_P}=4,\qquad\frac{\lambda_Q}{\lambda_P}=2} \]

Explanation: \(\textbf{Calculate the photon wavelength:}\) Photon energy is: \[ E_\gamma=\frac{hc}{\lambda_\gamma} \] Therefore: \[ \lambda_\gamma=\frac{hc}{E_\gamma} \] Substituting: \[ \lambda_\gamma=\frac{1240\,\text{eV nm}}{4.0\,\text{eV}} \] \[ \lambda_\gamma=310\,\text{nm} \] \(\textbf{Determine the electron kinetic energy in joules:}\) The electron kinetic energy is: \[ K_e=4.0\,\text{eV} \] Thus: \[ K_e=4.0(1.60\times10^{-19})\,\text{J} \] \[ K_e=6.40\times10^{-19}\,\text{J} \] \(\textbf{Recover electron momentum:}\) For non-relativistic motion: \[ K_e=\frac{p_e^2}{2m_e} \] Hence: \[ p_e=\sqrt{2m_eK_e} \] Substituting: \[ p_e=\sqrt{2(9.10\times10^{-31})(6.40\times10^{-19})} \] \[ p_e\approx1.08\times10^{-24}\,\text{kg m s}^{-1} \] \(\textbf{Calculate the electron matter wavelength:}\) Using: \[ \lambda_e=\frac{h}{p_e} \] we obtain: \[ \lambda_e=\frac{6.63\times10^{-34}}{1.08\times10^{-24}}\,\text{m} \] \[ \lambda_e\approx6.14\times10^{-10}\,\text{m} \] \[ \lambda_e\approx0.614\,\text{nm} \] \(\textbf{Compare momenta through the wavelengths:}\) Both photon and electron satisfy: \[ p=\frac{h}{\lambda} \] Therefore, the particle having the shorter wavelength has the greater momentum magnitude. Here: \[ \lambda_e\ll\lambda_\gamma \] so: \[ p_e\gg p_\gamma \] \(\textbf{Physical interpretation:}\) The photon and electron have equal numerical energies but obey different energy-momentum relations. Photon energy is \(E=pc\), while the non-relativistic electron kinetic energy is \(K=p^2/(2m_e)\). Their momenta and associated wavelengths are therefore very different despite the equality of the specified energies. \(\textbf{Final answer:}\) \[ \boxed{\lambda_\gamma=310\,\text{nm},\qquad\lambda_e\approx0.614\,\text{nm}} \]

Explanation: Observation P is consistent because de Broglie's relation states that wavelength is inversely proportional to momentum magnitude. Doubling momentum therefore reduces wavelength to half its original value. Observation Q is also consistent when the electron remains non-relativistic. In that regime, momentum magnitude is proportional to the square root of kinetic energy for an unchanged particle mass. Increasing kinetic energy by a factor of four therefore doubles momentum and halves the wavelength. Observation R is incorrect because reversing a momentum vector does not, by itself, change its magnitude. De Broglie wavelength depends on momentum magnitude rather than the sign of its component along an arbitrarily chosen axis. An ideal elastic rebound can reverse the direction of motion while preserving kinetic energy, speed and momentum magnitude. The wavelength magnitude then remains unchanged. These observations distinguish three different changes: changing momentum magnitude directly, changing it indirectly through kinetic energy and changing momentum direction alone. Only the first two necessarily alter the wavelength in the stated manner.

Explanation: \(\textbf{Determine the two photoelectron energies:}\) The incident photon energy is: \[ E_\gamma=5.0\,\text{eV} \] For metal P: \[ K_P=E_\gamma-\phi_P \] \[ K_P=5.0-2.0 \] \[ K_P=3.0\,\text{eV} \] For metal Q: \[ K_Q=5.0-3.0 \] \[ K_Q=2.0\,\text{eV} \] \(\textbf{Express the condition for equal matter wavelengths:}\) Both emitted particles are electrons and therefore have equal masses. Their de Broglie wavelengths are: \[ \lambda=\frac{h}{\sqrt{2m_eK}} \] Consequently, equal wavelengths require equal kinetic energies. \(\textbf{Find the required acceleration for Q:}\) P enters the diffraction apparatus directly with: \[ K_{P,\text{apparatus}}=3.0\,\text{eV} \] Q gains electrical energy: \[ \Delta K_Q=eV \] Therefore: \[ K_{Q,\text{apparatus}}=2.0\,\text{eV}+eV \] For wavelength equality: \[ 2.0\,\text{eV}+eV=3.0\,\text{eV} \] Hence: \[ eV=1.0\,\text{eV} \] \[ V=1.0\,\text{V} \] \(\textbf{Consider a later common acceleration:}\) Once the beams have equal kinetic energies, a further equal kinetic-energy gain preserves that equality. If each beam subsequently gains \(eU\): \[ K'_{P}=3.0\,\text{eV}+eU \] \[ K'_{Q}=3.0\,\text{eV}+eU \] Thus: \[ K'_P=K'_Q \] Their momentum magnitudes and de Broglie wavelengths remain equal. \(\textbf{Physical interpretation:}\) Different photoelectric work functions initially produce different electron kinetic energies under the same photon illumination. A suitable electrical acceleration can compensate for that difference. Once equal kinetic energies have been established for equal-mass particles, an identical additional energy change preserves their equal matter wavelengths. \(\textbf{Final answer:}\) \[ \boxed{V=1.0\,\text{V},\qquad\lambda'_P=\lambda'_Q\ \text{after equal subsequent acceleration}} \]

Explanation: \(\textbf{Relate molecular temperature to momentum:}\) For a gas particle of mass \(m\), the root-mean-square speed is: \[ v_{\mathrm{rms}}=\sqrt{\frac{3kT}{m}} \] The momentum magnitude corresponding to this speed is: \[ p_{\mathrm{rms}}=mv_{\mathrm{rms}} \] Substituting: \[ p_{\mathrm{rms}}=m\sqrt{\frac{3kT}{m}} \] \[ p_{\mathrm{rms}}=\sqrt{3mkT} \] \(\textbf{Construct the corresponding matter wavelength:}\) Using the de Broglie relation: \[ \lambda=\frac{h}{p} \] we obtain: \[ \lambda_{\mathrm{rms}}=\frac{h}{\sqrt{3mkT}} \] Thus: \[ \lambda_{\mathrm{rms}}\propto\frac{1}{\sqrt{mT}} \] \(\textbf{Compare the samples initially:}\) Both samples begin at temperature \(T\). The mass ratio is: \[ \frac{m_N}{m_{\mathrm{He}}}=\frac{28u}{4u}=7 \] Therefore: \[ \frac{\lambda_{\mathrm{He}}}{\lambda_N}=\sqrt{\frac{m_N}{m_{\mathrm{He}}}} \] \[ \frac{\lambda_{\mathrm{He}}}{\lambda_N}=\sqrt{7} \] Helium initially has the larger wavelength associated with its root-mean-square speed. \(\textbf{Determine the required helium temperature:}\) Let the final helium temperature be \(T'\). Equal wavelengths require: \[ \sqrt{m_{\mathrm{He}}T'}=\sqrt{m_NT} \] Squaring: \[ m_{\mathrm{He}}T'=m_NT \] Therefore: \[ T'=\frac{m_N}{m_{\mathrm{He}}}T \] \[ T'=7T \] \(\textbf{Physical interpretation:}\) At equal temperatures, the lighter helium atoms have smaller root-mean-square momenta and hence larger associated wavelengths. Heating helium increases its characteristic momentum. The required temperature increase compensates exactly for the mass difference. The calculation concerns the wavelength corresponding to root-mean-square speed; molecules in a thermal gas have a distribution of speeds and therefore a distribution of individual de Broglie wavelengths. \(\textbf{Final answer:}\) \[ \boxed{T'=7T} \]