201. Rutherford's nuclear model describes an atom as having
ⓐ. a uniformly positive sphere with electrons embedded in it
ⓑ. only neutrons spread throughout the atom
ⓒ. electrons fixed inside the nucleus with no empty space
ⓓ. small positive nucleus with revolving electrons
Correct Answer: small positive nucleus with revolving electrons
Explanation: Rutherford proposed that most of the atom is empty space and that positive charge is concentrated in a small central nucleus. Electrons were imagined to revolve around this nucleus. This model explained why most alpha particles passed straight through the foil and why a few were strongly deflected. It was different from Thomson's model, where positive charge was spread throughout the atom. The model introduced the nuclear picture of the atom, but it did not yet explain electron stability or line spectra.
202. In Rutherford's model, electrons do not simply fly away from the nucleus because the model assumes an attractive force between
ⓐ. electrons and the positive nucleus
ⓑ. neutral neutrons and photons
ⓒ. two negatively charged electrons only
ⓓ. alpha particles and the fluorescent screen
Correct Answer: electrons and the positive nucleus
Explanation: Rutherford's model places a positively charged nucleus at the centre of the atom. Electrons are negatively charged, so they are attracted toward the positive nucleus by electrostatic attraction. This attraction was used to picture electrons revolving around the nucleus. The attraction is not between neutrons and photons, and it is not caused by the fluorescent screen used in the scattering experiment. The model depends on opposite charges attracting each other within the atom.
203. The part of Rutherford's model that most directly explains the passage of most alpha particles through gold foil is that
ⓐ. atoms are mostly empty space
ⓑ. electrons have no charge
ⓒ. the nucleus occupies nearly the whole atom
ⓓ. positive charge is spread uniformly throughout the atom
Correct Answer: atoms are mostly empty space
Explanation: Rutherford observed that most alpha particles passed through the foil without appreciable deflection. This could be explained if most of the atom's volume contained empty space. Only particles passing very close to the tiny positive nucleus experienced strong repulsion. If the nucleus occupied nearly the whole atom, most alpha particles would have been deflected strongly. The conclusion of empty space is therefore tied to the high number of undeflected alpha particles.
204. Two statements about Rutherford's nuclear model are given.
Statement I: It explains the presence of a small, dense, positively charged nucleus.
Statement II: It fully explains why revolving electrons do not lose energy.
ⓐ. Both Statement I and Statement II are true
ⓑ. Statement I is true, but Statement II is false
ⓒ. Statement I is false, but Statement II is true
ⓓ. Both Statement I and Statement II are false
Correct Answer: Statement I is true, but Statement II is false
Explanation: Rutherford's model successfully introduced the idea of a tiny, dense, positively charged nucleus. This idea was strongly supported by alpha-particle scattering observations. However, the model pictured electrons revolving around the nucleus without explaining why they would remain stable. According to classical electromagnetic theory, an accelerating charged particle should lose energy by radiation. Statement II therefore describes a limitation that Rutherford's model could not solve.
205. The classical stability problem in Rutherford's model arises because a revolving electron is a charged particle that should
ⓐ. become a neutron immediately in the foil experiment
ⓑ. increase its mass number continuously
ⓒ. stop interacting with the nucleus in the foil experiment
ⓓ. radiate energy and spiral into the nucleus
Correct Answer: radiate energy and spiral into the nucleus
Explanation: In Rutherford's model, electrons revolve around the nucleus. A revolving electron is undergoing acceleration because its direction of motion changes continuously. Classical electromagnetic theory predicts that an accelerating charged particle should emit radiation and lose energy. If the electron loses energy, it should move closer and closer to the nucleus and finally collapse into it. Since real atoms are stable, this prediction revealed a serious limitation of Rutherford's model.
206. Assertion: Rutherford's model could not explain the stability of atoms.
Reason: A revolving electron should continuously lose energy according to classical theory.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Rutherford's model treated electrons as particles revolving around the nucleus. Such electrons would be accelerating because circular motion involves continuous change in direction. Classical theory says an accelerating charged particle should emit radiation and lose energy. This would make the electron spiral into the nucleus, making the atom unstable. The Reason directly explains why Rutherford's model failed to account for atomic stability.
207. A model predicts that an electron moving around the nucleus should emit energy continuously. If this were true, the atom would be expected to show
ⓐ. perfect stability with no change in electron motion
ⓑ. no interaction between nucleus and electron
ⓒ. collapse of the electron into the nucleus
ⓓ. a nucleus with negative charge
Correct Answer: collapse of the electron into the nucleus
Explanation: Continuous emission of energy would reduce the energy of the revolving electron. With less energy, the electron would not remain in the same orbit around the nucleus. It would spiral inward and eventually fall into the positive nucleus. This is not observed for ordinary atoms, which are stable under normal conditions. The prediction shows why the classical orbit picture was incomplete.
208. Rutherford's model also failed to explain the line spectrum of hydrogen because it did not include
ⓐ. protons in the nucleus in the foil experiment
ⓑ. quantized electron energy levels
ⓒ. empty space inside the atom in the foil experiment
ⓓ. positive charge in the nucleus
Correct Answer: quantized electron energy levels
Explanation: A line spectrum contains radiation of only certain definite frequencies. Rutherford's model did not provide the idea that electrons can have only certain allowed energies. Without quantized energy levels, the model could not explain why atoms emit or absorb radiation in discrete lines. The model did include a positive nucleus and empty space, so those are not the missing features in this context. Explaining line spectra required a new idea about allowed electron energies.
209. A comparison of model features is shown below.
| Row | Feature | Best linked model |
|---|
| P | Positive charge spread throughout atom | Thomson model |
| Q | Tiny dense positive nucleus | Rutherford model |
| R | Classical stability of revolving electrons fully explained | Rutherford model |
| S | Atom mostly empty space | Rutherford model |
The mismatched row is
ⓐ. R
ⓑ. P
ⓒ. Q
ⓓ. S
Correct Answer: R
Explanation: Thomson's model treated positive charge as spread throughout the atom, so row \(P\) is matched correctly. Rutherford's model introduced the small dense positive nucleus and explained that the atom is mostly empty space, so rows \(Q\) and \(S\) are also correct. Row \(R\) is mismatched because Rutherford's model did not fully explain the stability of revolving electrons. The classical stability problem was one of the main reasons a quantized model was later needed.
210. Use the arrangement described below. An electron is pictured as moving in a circular path around a positive nucleus, just as Rutherford's model imagined. Classical electromagnetic theory is then applied to the moving electron. The expected difficulty is that the electron should
ⓐ. keep exactly the same energy forever
ⓑ. become positively charged at once in the foil experiment
ⓒ. remove all protons from the nucleus in the foil experiment
ⓓ. radiate energy as an accelerating charged particle
Correct Answer: radiate energy as an accelerating charged particle
Explanation: Circular motion involves acceleration because the direction of velocity changes continuously. An electron is a charged particle, so classical electromagnetic theory predicts that it should radiate energy when accelerated. This energy loss would make the electron move closer to the nucleus. The prediction contradicts the observed stability of atoms. The weakness is not that the electron becomes positive, but that the classical orbit picture cannot keep the electron stable.
211. The need for a model beyond Rutherford's nuclear model arose mainly because Rutherford's model could not explain
ⓐ. atomic stability and line spectra
ⓑ. the presence of a nucleus
ⓒ. why most alpha particles passed through the atom
ⓓ. the positive charge of a proton
Correct Answer: atomic stability and line spectra
Explanation: Rutherford's model explained the nuclear nature of the atom and the large empty space inside it. It also explained why some alpha particles were deflected strongly. However, it could not explain why revolving electrons did not radiate energy and collapse into the nucleus. It also could not explain why atomic spectra contain definite lines instead of a continuous spread of frequencies. These limitations prepared the way for models using quantized electron energies.
212. Electromagnetic radiation is best described as energy travelling through space with
ⓐ. only stationary protons in space in the notation
ⓑ. only moving neutrons in matter in the notation
ⓒ. nuclei arranged in a fixed solid line
ⓓ. oscillating electric and magnetic fields
Correct Answer: oscillating electric and magnetic fields
Explanation: Electromagnetic radiation consists of oscillating electric and magnetic fields. These fields are associated with the propagation of energy through space. Light is a familiar form of electromagnetic radiation. This description is different from a beam of material particles such as electrons, protons, or neutrons. The wave picture of radiation becomes important for understanding wavelength, frequency, velocity, and later atomic spectra.
213. In the wave description of electromagnetic radiation, wavelength \(\lambda\) refers to
ⓐ. the number of waves passing a point per second
ⓑ. the charge carried by one photon
ⓒ. distance between successive in-phase points
ⓓ. the number of protons in the nucleus
Correct Answer: distance between successive in-phase points
Explanation: Wavelength is a distance measurement in a wave. It can be understood as the distance between two successive crests, two successive troughs, or any two successive points in the same phase. Frequency is different; it measures how many wave cycles pass a point per second. Wavelength is commonly represented by \(\lambda\). Confusing wavelength with frequency reverses the meaning of two basic wave quantities.
214. Frequency \(\nu\) of electromagnetic radiation means the
ⓐ. wave cycles passing a point per second
ⓑ. distance between two successive crests
ⓒ. total number of neutrons in an atom
ⓓ. charge-to-mass ratio of an electron
Correct Answer: wave cycles passing a point per second
Explanation: Frequency tells how many complete wave cycles pass a point in one second. It is represented by \(\nu\) and is measured in \(\text{s}^{-1}\) or \(\text{Hz}\). Wavelength \(\lambda\) is a distance, not a count per second. Frequency and wavelength are related for electromagnetic radiation through the wave equation \(c=\lambda\nu\). Keeping time-related and distance-related meanings separate is essential before using the formula.
215. The speed of electromagnetic radiation in vacuum is represented by
ⓐ. \(m_e\)
ⓑ. \(Z\)
ⓒ. \(c\)
ⓓ. \(A\)
Correct Answer: \(c\)
Explanation: The symbol \(c\) represents the speed of electromagnetic radiation in vacuum. For light and other electromagnetic waves in vacuum, this speed is a constant. The symbols \(Z\) and \(A\) belong to atomic number and mass number, not wave motion. The symbol \(m_e\) represents electron mass. The wave symbols \(\lambda\), \(\nu\), and \(c\) should be kept distinct from nuclear and particle symbols.
216. The wave relation for electromagnetic radiation in vacuum is
ⓐ. \(c=\frac{\nu}{\lambda}\)
ⓑ. \(Z=A+N\)
ⓒ. \(c=\lambda\nu\)
ⓓ. \(m_e=e\nu\)
Correct Answer: \(c=\lambda\nu\)
Explanation: Electromagnetic radiation in vacuum follows the relation \(c=\lambda\nu\). Here \(c\) is the velocity of radiation, \(\lambda\) is wavelength, and \(\nu\) is frequency. The relation shows that wavelength and frequency are inversely related when \(c\) is constant. The nuclear symbols \(Z\), \(A\), and \(N\) belong to atomic composition and not to wave speed. The formula must be read as speed equals wavelength multiplied by frequency.
217. If the wavelength of electromagnetic radiation increases while it travels in vacuum, its frequency must
ⓐ. increase
ⓑ. decrease
ⓒ. remain fixed for all wavelengths
ⓓ. become equal to atomic number
Correct Answer: decrease
Explanation: In vacuum, electromagnetic radiation travels with constant speed \(c\). The relation \(c=\lambda\nu\) connects wavelength and frequency. If \(\lambda\) increases while \(c\) remains constant, \(\nu\) must decrease to keep the product \(\lambda\nu\) equal to \(c\). This is an inverse relation, not a direct relation. A longer wavelength therefore corresponds to a lower frequency for radiation travelling in vacuum.
218. A wave has wavelength \(3.0\times10^{-7}\,\text{m}\) in vacuum. Taking \(c=3.0\times10^8\,\text{m s}^{-1}\), its frequency is
ⓐ. \(1.0\times10^{-15}\,\text{s}^{-1}\)
ⓑ. \(9.0\times10^1\,\text{s}^{-1}\)
ⓒ. \(9.0\times10^{-1}\,\text{s}^{-1}\)
ⓓ. \(1.0\times10^{15}\,\text{s}^{-1}\)
Correct Answer: \(1.0\times10^{15}\,\text{s}^{-1}\)
Explanation: \( \textbf{Given data:} \) Wavelength \(\lambda=3.0\times10^{-7}\,\text{m}\).
\( \textbf{Given data:} \) Speed \(c=3.0\times10^8\,\text{m s}^{-1}\).
\( \textbf{Required quantity:} \) Frequency \(\nu\).
\( \textbf{Wave relation:} \)
\[
c=\lambda\nu
\]
\( \textbf{Rearrange:} \)
\[
\nu=\frac{c}{\lambda}
\]
\( \textbf{Substitution:} \)
\[
\nu=\frac{3.0\times10^8}{3.0\times10^{-7}}\,\text{s}^{-1}
\]
\( \textbf{Number part:} \)
\[
\frac{3.0}{3.0}=1.0
\]
\( \textbf{Power of ten part:} \)
\[
10^8/10^{-7}=10^{15}
\]
\( \textbf{Final answer:} \) \(\nu=1.0\times10^{15}\,\text{s}^{-1}\). The metre unit cancels, leaving frequency in per second.
219. In a graph of frequency \(\nu\) on the y-axis against wavelength \(\lambda\) on the x-axis for electromagnetic radiation in vacuum, the curve should show
ⓐ. frequency increasing linearly with wavelength
ⓑ. frequency decreasing as wavelength increases
ⓒ. frequency remaining zero at all wavelengths
ⓓ. wavelength independent of frequency only at high values
Correct Answer: frequency decreasing as wavelength increases
Explanation: For electromagnetic radiation in vacuum, \(c=\lambda\nu\). Since \(c\) is constant, frequency is inversely proportional to wavelength. As wavelength increases, frequency decreases. The graph of \(\nu\) against \(\lambda\) is therefore a decreasing curve, not a straight line with positive slope. A direct-line graph would incorrectly treat \(\lambda\) and \(\nu\) as directly proportional.
220. A table lists wave quantities for electromagnetic radiation.
| Row | Quantity | Meaning |
|---|
| P | \(\lambda\) | distance between successive points in the same phase |
| Q | \(\nu\) | number of wave cycles per second |
| R | \(c\) | speed of radiation in vacuum |
| S | \(\lambda\) | number of protons in the nucleus |
The mismatched row is
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: The symbol \(\lambda\) represents wavelength, a distance between successive points in the same phase of a wave. The symbol \(\nu\) represents frequency, the number of cycles per second. The symbol \(c\) represents the speed of electromagnetic radiation in vacuum. Row \(S\) incorrectly connects \(\lambda\) with proton number, which is represented by \(Z\) in atomic structure. Wave symbols and nuclear symbols must be kept separate because they represent different physical quantities.
