Physical Chemistry Questions and Answers – Calculation of the Energy of a Hydrogen Molecule Ion

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

Answer: b
Explanation: The formula for the energy required to remove an electron is
En = \(\frac{-13.6}{n^2}\)eV
Therefore, energy required to remove an electron from n = 2 is
E = E2 = 0 + 13.6/22 = 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

Answer: d
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) 9r0
b) r0
c) 3r0
d) r0/9
View Answer

Answer: a
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 rn = rnn2 is used. Therefore, r3 = 9r0 is the radius of the third orbit in Bohr’s model.
advertisement
advertisement

4. What is the energy corresponding to a transition between 3rd and 4th 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

Answer: d
Explanation: The formula for the energy required to remove an electron is
En = \(\frac{-13.6}{n^2}\)eV
Therefore, energy required to a transition between 3rd and 4th orbits is
△E = E4 – E3 = -13.6/42 + 13.6/32 = 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

Answer: b
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

Answer: c
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

Answer: d
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.
advertisement

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

Answer: b
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 6th orbit is 3h/π.
a) True
b) False
View Answer

Answer: a
Explanation: The angular momentum in the nth orbit is given by, mvr=nh/2π. Therefore, for the angular momentum in 6th orbit put n = 6 in the given formula, so we get angular momentum of an electron in 6th orbital that is, 3h/π. So, the statement is true.
advertisement

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

Answer: a
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.

To practice all areas of Physical Chemistry, here is complete set of Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.