Rutherford's Model & Atomic Structure — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Rutherford's Model & Atomic Structure MCQs with step-by-step solutions (21 questions). Part of Structure of Atoms and Nuclei. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Rutherford's Model & Atomic Structure · easy · theory
Rutherford's alpha particle scattering experiment used a thin foil of:
A. Gold ✓ Correct
B. Iron
C. Lead
D. Aluminium
Solution: Gold can be beaten into extremely thin foil, only a few hundred atoms thick, so multiple scattering is negligible.
Q2 — Rutherford's Model & Atomic Structure · easy · theory
In the alpha scattering experiment, most of the alpha particles passed through the foil undeviated. This showed that:
A. Electrons are heavier than alpha particles
B. The nucleus is negatively charged
C. The atom has a uniform positive charge
D. Most of the atom is empty space ✓ Correct
Solution: Only a tiny fraction of the incident particles encountered anything massive enough to deflect them.
Q3 — Rutherford's Model & Atomic Structure · medium · theory
The fact that a very few alpha particles were deflected through large angles showed that:
A. The atom is a uniform sphere of positive charge
B. Electrons occupy the centre of the atom
C. Alpha particles are negatively charged
D. The positive charge and mass are concentrated in a very small nucleus ✓ Correct
Solution: Rutherford described it as being as surprising as a shell bouncing back from tissue paper.
Q4 — Rutherford's Model & Atomic Structure · easy · theory
The nucleus of an atom carries:
A. A negative charge
B. A positive charge ✓ Correct
C. A charge that varies with time
D. No charge
Solution: The protons within it give the nucleus a charge of $+Ze$.
Q5 — Rutherford's Model & Atomic Structure · medium · theory
The chief drawback of Rutherford's nuclear model was that:
A. It could not explain the scattering results
B. An orbiting electron should radiate energy and spiral into the nucleus ✓ Correct
C. It predicted that atoms have no mass
D. It required the nucleus to be negatively charged
Solution: Classical electromagnetism demands that an accelerated charge radiate, making the atom collapse in about $10^{-8}\text{ s}$.
Q6 — Rutherford's Model & Atomic Structure · medium · theory
Rutherford's model could not explain:
A. The large angle scattering of alpha particles
B. The discrete line spectra emitted by atoms ✓ Correct
C. The neutrality of the atom
D. The existence of the nucleus
Solution: A continuously spiralling electron would radiate a continuous spectrum, not sharp lines.
Q7 — Rutherford's Model & Atomic Structure · medium · theory
The distance of closest approach of an alpha particle to a nucleus is the distance at which:
A. It is captured by the nucleus
B. Its entire kinetic energy has been converted into electrostatic potential energy ✓ Correct
C. It begins to emit radiation
D. Its kinetic energy is maximum
Solution: At that point the particle is momentarily at rest before being repelled back.
Q8 — Rutherford's Model & Atomic Structure · hard · theory
In the alpha scattering experiment, the number of particles scattered through an angle $\theta$ varies as:
A. $\sin^4(\theta/2)$
B. $\dfrac{1}{\sin^2(\theta/2)}$
C. $\cos^2(\theta/2)$
D. $\dfrac{1}{\sin^4(\theta/2)}$ ✓ Correct
Solution: This steep dependence is the signature of Rutherford's inverse-square Coulomb scattering.
Q9 — Rutherford's Model & Atomic Structure · hard · numerical
The distance of closest approach of an alpha particle of kinetic energy $K$ to a nucleus of atomic number $Z$ is:
A. $\dfrac{2Ze^2K}{4\pi\varepsilon_0}$
B. $\dfrac{4\pi\varepsilon_0K}{2Ze^2}$
C. $\dfrac{Ze^2}{4\pi\varepsilon_0K}$
D. $\dfrac{2Ze^2}{4\pi\varepsilon_0K}$ ✓ Correct
Solution: Equating the initial kinetic energy to the Coulomb potential energy $\dfrac{(2e)(Ze)}{4\pi\varepsilon_0r_0}$ gives this result.
Q10 — Rutherford's Model & Atomic Structure · hard · numerical
An alpha particle of $5\text{ MeV}$ approaches a gold nucleus ($Z = 79$). Its distance of closest approach is approximately (taking $ke^2 = 1.44\text{ MeV}\cdot\text{fm}$):
A. $22.8\text{ fm}$
B. $45.5\text{ fm}$ ✓ Correct
C. $91\text{ fm}$
D. $4.55\text{ fm}$
Solution: $r_0 = \dfrac{2Zke^2}{K} = \dfrac{2 \times 79 \times 1.44}{5} \approx 45.5\text{ fm}$.
Q11 — Rutherford's Model & Atomic Structure · medium · numerical
If the kinetic energy of the bombarding alpha particle is doubled, the distance of closest approach:
A. Halves ✓ Correct
B. Remains unchanged
C. Becomes four times
D. Doubles
Solution: $r_0 \propto \dfrac{1}{K}$.
Q12 — Rutherford's Model & Atomic Structure · medium · numerical
If the atomic number of the target nucleus is doubled, the distance of closest approach:
A. Doubles ✓ Correct
B. Halves
C. Remains unchanged
D. Becomes four times
Solution: $r_0 \propto Z$ for a given bombarding energy.
Q13 — Rutherford's Model & Atomic Structure · medium · numerical
An alpha particle scattered through $180^\circ$ has undergone:
A. Capture by the nucleus
B. No interaction at all
C. A grazing collision with a large impact parameter
D. A head-on collision with zero impact parameter ✓ Correct
Solution: Only a direct head-on approach can reverse the particle completely.
Q14 — Rutherford's Model & Atomic Structure · hard · numerical
An alpha particle of $7.7\text{ MeV}$ approaches a gold nucleus ($Z = 79$). Its distance of closest approach is approximately:
A. $14.8\text{ fm}$
B. $59\text{ fm}$
C. $45.5\text{ fm}$
D. $29.5\text{ fm}$ ✓ Correct
Solution: $r_0 = \dfrac{2 \times 79 \times 1.44}{7.7} \approx 29.5\text{ fm}$.
Q15 — Rutherford's Model & Atomic Structure · medium · numerical
The radius of a nucleus is of the order of $10^{-15}\text{ m}$ and that of an atom $10^{-10}\text{ m}$. Their ratio is:
A. $10^{-10}$
B. $10^{-5}$ ✓ Correct
C. $10^5$
D. $10^{-25}$
Solution: The nucleus is about one hundred thousand times smaller in radius than the atom that contains it.
Q16 — Rutherford's Model & Atomic Structure · easy · numerical
The charge on an alpha particle is:
A. $+4e$
B. $+e$
C. $-2e$
D. $+2e$ ✓ Correct
Solution: An alpha particle is a helium nucleus, containing two protons and two neutrons.
Q17 — Rutherford's Model & Atomic Structure · easy · numerical
The mass number of an alpha particle is:
A. $2$
B. $4$ ✓ Correct
C. $8$
D. $1$
Solution: Two protons plus two neutrons give a mass number of four.
Q18 — Rutherford's Model & Atomic Structure · hard · numerical
Given that the nuclear radius is about $10^{-5}$ times the atomic radius, the fraction of the atomic volume occupied by the nucleus is of the order of:
A. $10^{-30}$
B. $10^{-10}$
C. $10^{-5}$
D. $10^{-15}$ ✓ Correct
Solution: Volume scales as the cube of the radius, so the ratio is $(10^{-5})^3 = 10^{-15}$.
Q19 — Rutherford's Model & Atomic Structure · medium · numerical
The distance of closest approach depends on the atomic number $Z$ and kinetic energy $K$ as:
A. $r_0 \propto \dfrac{K}{Z}$
B. $r_0 \propto \dfrac{1}{ZK}$
C. $r_0 \propto \dfrac{Z}{K}$ ✓ Correct
D. $r_0 \propto ZK$
Solution: A heavier target repels more strongly, while a faster projectile penetrates closer.
Q20 — Rutherford's Model & Atomic Structure · medium · numerical
An alpha particle with a large impact parameter is scattered through:
A. Exactly $90^\circ$
B. Exactly $180^\circ$
C. A small angle ✓ Correct
D. A large angle
Solution: Passing far from the nucleus, it experiences only a weak Coulomb repulsion and is barely deflected.
Q21 — Rutherford's Model & Atomic Structure · hard · numerical
In the alpha scattering experiment, the number of particles scattered at $90^\circ$ compared with those at $180^\circ$ is:
A. Zero
B. Smaller
C. Exactly equal
D. Greater, since scattering falls off rapidly with angle ✓ Correct
Solution: With $N \propto \dfrac{1}{\sin^4(\theta/2)}$, smaller angles are very much more probable.