This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Atomic Spectra”.

1. Which source is associated with a line emission spectrum?

a) Electric fire

b) Neon street sign

c) Red traffic light

d) Sun

View Answer

Explanation: Neon street sign gives a line emission spectrum. When neon atoms gain enough energy to become excited, light is produced. Atom releases a photon when it returns to a lower energy state. Therefore, the source associated with a line emission spectrum is the neon street sign.

2. According to the uncertainty principle for an electron, time measurement will become uncertain if which of the following is measured with high certainty?

a) Energy

b) Momentum

c) Location

d) Velocity

View Answer

Explanation: According to the uncertainty principle,

ΔE.Δt >= \(\frac {h}{2\pi }\).

Thus the time measured will become uncertain if ΔE is measured with high certainty.

3. Find the true statement.

a) An electron will not lose energy when jumping from the 1^{st} orbit to the 3^{rd} orbit

b) An electron will not give energy when jumping from the 1^{st} orbit to the 3^{rd} orbit

c) An electron will release energy when jumping from the 1^{st} orbit to the 3^{rd} orbit

d) An electron will absorb energy when jumping from the 1^{st} orbit to the 3^{rd} orbit

View Answer

Explanation: An electron will absorb energy when jumping from the 1

^{st}orbit to the 3

^{rd}orbit. Only by absorbing energy, an electron will be able to jump from the first orbit to the third orbit in the atomic spectrum.

4. The size of the atom is proportional to which of the following?

a) A

b) A^{1/3}

c) A^{2/3}

d) A^{-1/3}

View Answer

Explanation: The size of the atom is proportional to A

^{1/3}.

The electron affinity of the atom is inversely proportional to the atomic size. As the number of energy levels increases, the atomic size must increase.

5. Calculate the ratio of the kinetic energy for the n = 2 electron for the Li atom to that of Be^{+} ion?

a) \(\frac {9}{16}\)

b) \(\frac {3}{4}\)

c) 1

d) \(\frac {1}{2}\)

View Answer

Explanation: \(\frac {KE_{Li}}{KE_{Be}} = \big [ \frac {(\frac {Z_{Li}}{2} )}{(\frac {Z_{Be}}{2})} \big ]^2 \)

\(\frac {KE_{Li}}{KE_{Be}} = \big [ \frac {(\frac {3}{2} )}{(\frac {4}{2})} \big ]^2 \)

\(\frac {KE_{Li}}{KE_{Be}} = \frac {9}{16}\).

6. The Bohr model of atoms uses Einstein’s photoelectric equation.

a) True

b) False

View Answer

Explanation: Bohr model assumes that the angular momentum of electrons is quantized. Therefore, the Bohr model of the atoms involved is independent of Einstein’s photoelectric equation.

7. What is the ratio of minimum to maximum wavelength in the Balmer series?

a) 5:9

b) 5:36

c) 1:4

d) 3:4

View Answer

Explanation: For a wavelength of Balmer series,

\(\frac {1}{\lambda }\) = R\( [ \frac {1}{4} – \frac {1}{9} ] \)

\(\frac {1}{\lambda } = \frac {5R}{30}\).

\(\frac {\lambda_{min}}{\lambda_{max}} = \frac {5R}{36} \times \frac {4}{R} \)

\(\frac {\lambda_{min}}{\lambda_{max}} = \frac {5}{9}\)

8. The energy of characteristic X-ray is a consequence of which of the following?

a) The energy of the projectile electron

b) The thermal energy of the target

c) Transition in target atoms

d) Temperature

View Answer

Explanation: The energy of characteristic X-ray is a consequence of transition in target atoms. They cause emission or absorption of electromagnetic radiation. The other options are not responsible for the energy of characteristic X rays.

9. Find out the minimum energy required to take out the only one electron from the ground state of Li^{+}?

a) 13.6 eV

b) 122.4 eV

c) 25.3 eV

d) 67.9 eV

View Answer

Explanation: Ionization energy is given as:

E = 13.6 Z

^{2}eV

For Li

^{+}, Z = 3

E = 13.6 × 9

E = 122.4 eV

10. What is the energy required to ionize an H-atom from the third excited state, if ground state ionization energy of H-atom is 13.6 eV?

a) 1.5 eV

b) 3.4 eV

c) 13.6 eV

d) 12.1 eV

View Answer

Explanation: From third excited state, E = \(\frac {-13.6}{16}\)

E = -0.85 eV

Energy required to ionize H-atom from second excited state = 0 – (-0.85)

E = +0.85 eV

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