This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Calculation of the Energy of a Hydrogen Molecule Ion”.

1. How much energy is required to remove an electron from n = 2 orbital, if 13.6 eV energy is required to ionize the hydrogen atom?

a) 10.2 eV

b) 3.4 eV

c) 0 eV

d) 6.8 eV

View Answer

Explanation: The formula for the energy required to remove an electron is

E

_{n}= \(\frac{-13.6}{n^2}\)eV

Therefore, energy required to remove an electron from n = 2 is

E

_{∞}= E

_{2}= 0 + 13.6/2

^{2}= 3.4 eV.

2. Which of the following is an integral multiple of h/2π in Bohr’s model of an atom?

a) Angular momentum

b) Kinetic energy

c) Radius of an atom

d) Potential energy

View Answer

Explanation: Angular momentum is an integral multiple of h/2π in Bohr’s model of an atom. Angular momentum of an electron by Bohr is given by mvr or nh/2π where, v is the velocity, n is the orbit in which electron is, m is the mass of the electron and r is the radius of the electron.

3. What is the radius of the third orbit, if the atomic radius of the first orbit is rQ in Bohr’s model?

a) 9r_{0}

b) r_{0}

c) 3r_{0}

d) r_{0}/9

View Answer

Explanation: The atomic radius of the first orbit in Bohr’s model is rQ. So, to determine the radius of the third orbit in Bohr’s model the following formula r

_{n}= r

_{n}n

^{2}is used. Therefore, r

_{3}= 9r

_{0}is the radius of the third orbit in Bohr’s model.

4. What is the energy corresponding to a transition between 3^{rd} and 4^{th} orbits if the ionization energy of hydrogen atom is 13.6 eV?

a) 3.40 eV

b) 1.51 eV

c) 0.85 eV

d) 0.66 eV

View Answer

Explanation: The formula for the energy required to remove an electron is

E

_{n}= \(\frac{-13.6}{n^2}\)eV

Therefore, energy required to a transition between 3rd and 4th orbits is

△E = E

_{4}– E

_{3}= -13.6/4

^{2}+ 13.6/3

^{2}= 0.66 eV.

5. Which of the following spectral series in hydrogen atom gives spectral line of 4860 Å?

a) Lyman

b) Balmer

c) Paschen

d) Brackett

View Answer

Explanation: The Balmer series is one of a set of six named series describing the spectral line emissions of the hydrogen atom. Since spectral line of wavelength 4860 Å lies in the visible region of the spectrum which is Balmer series of the spectrum.

6. Which of the following spectral series falls within the visible range of electromagnetic radiation?

a) Pfund series

b) Paschen series

c) Balmer series

d) Lyman series

View Answer

Explanation: Balmer spectral series falls within the visible range of electromagnetic radiation. Lyman series of hydrogen atom lies in ultraviolet region. Paschen series and Pfund series lie in the infrared spectral series.

7. Which of the following will be the longest wavelength in Balmer series of hydrogen atom?

a) 2300 Å

b) 1215 Å

c) 4800 Å

d) 6557 Å

View Answer

Explanation: The Balmer series of spectral lines occur when electrons transition from an energy level higher than n = 2 back down to n = 1. Therefore, the longest wavelength in Balmer series is n1 = 2 and n2= 3 is 1/π=\(R[\frac{1}{n1^2}-\frac{1}{n2^2}]\),R=1.1×10 m

^{-1}. Thus, 6557 Å will be the longest wavelength.

8. Which of the following postulates of the Bohr’s model led to the quantization of energy of the hydrogen atom?

a) The electron goes around the nucleus in circular orbits

b) The angular momentum of the electron can only be an integral multiple of h/2π

c) The magnitude of the linear momentum of the electron is quantized

d) Quantization of energy is itself a postulate of the Bohr model

View Answer

Explanation: According to Bohr’s atomic model, the angular momentum of electron orbiting around the nucleus is quantized. The angular momentum of the electron is given as nh/2π, where n is an integer. Therefore, angular momentum of an electron can only be an integral multiple of h/2π.

9. According to the Bohr’s theory angular momentum of an electron in 6^{th} orbit is 3h/π.

a) True

b) False

View Answer

Explanation: The angular momentum in the n

^{th}orbit is given by, mvr=nh/2π. Therefore, for the angular momentum in 6

^{th}orbit put n = 6 in the given formula, so we get angular momentum of an electron in 6

^{th}orbital that is, 3h/π. So, the statement is true.

10. Is the following statement true about the Lyman spectral series?

Statement Lyman series lies in the ultraviolet region

a) Statement is true

b) Statement is false

View Answer

Explanation: The Lyman series lie in the ultraviolet region because when the hydrogen electron becomes excited to a higher energy level and then falls to the ground state, if it falls to n=1 energy level, then a photon is emitted by the electron. This transition from a higher energy level to n=1 is part of the Lyman series.

**Sanfoundry Global Education & Learning Series – Physical Chemistry.**

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