Physics Questions and Answers – Alpha-Particle Scattering and Rutherford’s Nuclear Model of Atom

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This set of Physics Online Quiz for Class 12 focuses on “Alpha-Particle Scattering and Rutherford’s Nuclear Model of Atom”.

1. The atomic number of silicon is 14. Its ground state electronic configuration is
a) 1s22s22p63s23p4
b) 1s22s22p63s23p3
c) 1s22s22p63s23p2
d) 1s22s22p63s23p1
View Answer

Answer: c
Explanation: The atomic number of silicon is 14.
Therefore, 14Si will have the following electronic configuration:
1s22s22p63s23p2
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2. What is the valence electron in alkali metal?
a) f-electron
b) p-electron
c) s-electron
d) d-electron
View Answer

Answer: c
Explanation: The valence electron in an alkali metal is an s-electron. Generally, they make up Group 1 of the periodic table. The different examples that come under this category are lithium, potassium, and francium.

3. Of the following pairs of species which one will have the same electronic configuration for both members?
a) Li+ and Na+
b) He and Ne+
c) H and Li
d) C and N+
View Answer

Answer: d
Explanation: Carbon and the positive ion of nitrogen (N+) will have the same electronic configuration.
The electronic configuration of both Carbon and the positive ion of nitrogen is as follows:
1s22s22p6.
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4. Which of the following did Bohr use to explain his theory?
a) Conservation of linear momentum
b) The quantization of angular momentum
c) Conservation of quantum frequency
d) Conservation of mass
View Answer

Answer: b
Explanation: To explain his theory, Niels Bohr used the quantization of angular momentum. It means the radius of the orbit and the energy will be quantized. The Boundary conditions for the wave function are periodic.

5. According to Bohr’s principle, what is the relation between the principal quantum number and the radius of the orbit?
a) r proportional to \(\frac {1}{n}\)
b) r proportional to \(\frac {1}{n^2}\)
c) r proportional to n
d) r proportional to n2
View Answer

Answer: d
Explanation: The equation is given as:
r = \(\frac {n^2 h^2}{4\pi^2 m k Z e^2}\)
Therefore, we can say that the radius of the orbit is directly proportional to the square of the principal quantum number.
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6. The kinetic energy of the α-particle incident on the gold foil is doubled. The distance of closest approach will also be doubled.
a) True
b) False
View Answer

Answer: b
Explanation: As the distance of the closest approach is inversely proportional to the kinetic energy of the incident α-particle, so the distance of the closest approach is halved when the kinetic energy of α-particle is doubled.

7. Based on the Bohr model, what is the minimum energy required to remove an electron from the ground state of Be atom? (Given: Z = 4)
a) 1.63 eV
b) 15.87 eV
c) 30.9 eV
d) 217.6 eV
View Answer

Answer: d
Explanation: The equation is given as:
En = \(\frac {-13.6 Z^2}{n^2}\) eV
En = \(\frac {-13.6 \times 16 }{ 1 }\)
En = -217.6 eV
Hence the ionization energy for an electron in the ground state of Be atom is 217.6 eV.
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8. If an α-particle collides head-on with a nucleus, what is its impact parameter?
a) Zero
b) Infinite
c) 10-10 m
d) 1010 m
View Answer

Answer: a
Explanation: The perpendicular distance between the path of a projectile and the center of the potential field is the impact parameter. Therefore, for a head-on collision of the α-particle with a nucleus, the impact parameter is equal to zero.

9. In which of the following system, will the radius of the first orbit (n=1) be minimum?
a) Doubly ionized lithium
b) Singly ionized helium
c) Deuterium atom
d) Hydrogen atom
View Answer

Answer: a
Explanation: The equation is given as:
r = \(\frac {h^2}{2\pi^2 m k Z e^2}\)
Hence, of the given atoms/ions, (Z = 3) is maximum for doubly ionized lithium, so the radius of its first orbit is minimum.
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10. An α-particle of energy 10 MeV is scattered through 180o by a fixed uranium nucleus. Calculate the order of distance of the closest approach?
a) 10-20cm
b) 10-12cm
c) 10-11cm
d) 1012cm
View Answer

Answer: b
Explanation: r0 = \(\frac {(2Ze^2)}{(4\pi \varepsilon_0 (\frac {1}{2}m v^2))}\)
r0 = \(\frac {9 \times 10^9 \times 2 \times 92 \times (1.6 \times 10^{-19})^2}{10 \times 1 \times 10^{-13}}\) J
r0 = 4.239 × 10-14 m
r0 = 4.2 × 10-12 cm.

Sanfoundry Global Education & Learning Series – Physics – Class 12.

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