Physical Chemistry Questions and Answers – Schrodinger Equation for Hydrogen Like Atoms

This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Schrodinger Equation for Hydrogen Like Atoms”.

1. In a two-particle system consisting of an electron and a nucleus, how does the potential energy of the system change if the radius between the electron and nucleus increased by a factor of two?
a) Decrease by a factor of two
b) Increases by a factor of two
c) Stays the same
d) Decrease by a factor of four
View Answer

Answer: a
Explanation: Potential energy V is given by V = \(\frac{-Ze^2}{4\pi \varepsilon_0r}\). Increasing the radius by a factor of two decreases the potential energy by a factor of two from this expression. This is because V is inversely proportional to r.

2. If the mass of an electron is me and mass of the nucleus is mn, what is the reduced mass μ of the electron-nucleus system?
a) μ = \(\frac{m_e+m_n}{2}\)
b) μ = \(\frac{m_em_n}{2}\)
c) μ = \(\frac{m_e+m_n}{m_em_n}\)
d) μ = \(\frac{m_em_n}{m_e+m_n}\)
View Answer

Answer: d
Explanation: Reduced mass is defined as the product of masses divided by their sum. In a way, it is a weighted mass average of a system consisting of more than one particle. Hence the reduced mass of a two-particle system consisting of a nucleus and an electron is given by μ = \(\frac{m_em_n}{m_e+m_n}\)

3. If the atomic number Z of a nucleus is increased by a factor of four, how does it affect the absolute value of energy levels of a hydrogen like atomic system.
a) Increases by a factor of 16
b) Increases by a factor of 4
c) Decreases by a factor of 16
d) Increases by a factor of 8
View Answer

Answer: a
Explanation: Energy levels for a hydrogen like atom are given by E = –\(\frac{\mu e^4 Z^2}{2(4\pi \varepsilon_0)^2 \hbar^2 n^2}\). Increasing the atomic number by a factor of four increases the energy by a factor of sixteen from this expression. This is because E is directly proportional to the square of Z.
advertisement
advertisement

5. If the atomic reduced mass of an electron-nucleus is increased by a factor of four, how does it affect the absolute value of energy levels of a hydrogen like atomic system.
a) Increases by a factor of 16
b) Increases by a factor of 4
c) Decreases by a factor of 16
d) Increases by a factor of 8
View Answer

Answer: b
Explanation: Energy levels for a hydrogen like atom are given by E = –\(\frac{\mu e^4 Z^2}{2(4\pi \varepsilon_0)^2 \hbar^2 n^2}\). Increasing the reduced mass by a factor of four increases the energy by a factor of four from this expression. This is because E is directly proportional to the reduced mass.

6. If the energy level of a nucleus is increased by a factor of two, how does it affect the energy levels of a hydrogen like atom system.
a) Increases by a factor of 2
b) Increases by a factor of 4
c) Decreases by a factor of 4
d) Increases by a factor of 2
View Answer

Answer: c
Explanation: Energy levels for a hydrogen like atom are given by E = –\(\frac{\mu e^4 Z^2}{2(4\pi \varepsilon_0)^2 \hbar^2 n^2}\). Increasing the energy level n by a factor of two increases the energy by a factor of four from this expression. This is because E is inversely proportional to the square of n.

7. What is the value of Bohr radius a0 = \(\frac{h^2 \varepsilon_0}{\pi m_e e^2}\) in picometers (pm)?
a) 203.32 pm
b) 12.45 pm
c) 52.92 pm
d) 103.84 pm
View Answer

Answer: c
Explanation: Evaluating all the constants in a0 = \(\frac{h^2 \varepsilon_0}{\pi m_e e^2}\) in terms of SI units gives Bohr’s radius as = 52.9177 × 10-12m. This constant is very important in quantum mechanics for evaluating energy levels after solving different types and forms of Schrodinger’s equation.

8. What is the value of Hartree energy Eh = \(\frac{e^2}{4\pi \varepsilon_0 a_0}\) in eV?
a) 27.21 eV
b) 103.43 eV
c) 90.99 eV
d) 132.24 eV
View Answer

Answer: a
Explanation: Plugging in the value of all physical constants in the expression Eh = \(\frac{e^2}{4\pi \varepsilon_0 a_0}\) gives Eh = 4.360 × 10-18J = 27.21 eV. This is a very important constant because the energy of a hydrogen atom in its ground state is equal to \(\frac{E_h}{2}\).
advertisement

9. The expression for Hartree energy is Eh = \(\frac{e^2}{4\pi \varepsilon_0 a_0}\). What is its expression purely in terms of fundamental constants, eliminating Bohr’s radius?
a) Eh = \(\frac{m_e e}{(4\pi \varepsilon_0)^2 \hbar}\)
b) Eh = \(\frac{m_e e^4}{\hbar^2}\)
c) Eh = \(\frac{m_e e^4 \hbar^2}{(4\pi \varepsilon_0)^2}\)
d) Eh = \(\frac{m_e e^4}{(4\pi \varepsilon_0)^2 \hbar^2}\)
View Answer

Answer: d
Explanation: Generally, Eh = \(\frac{e^2}{4\pi \varepsilon_0 a_0}\) and a0 = \(\frac{h^2 \varepsilon_0}{\pi m_e e^2}\). Plugging the value of Bohr’s radius in Hartree energy gives Eh = \(\frac{e^2 \pi m_e e^2}{4\pi \varepsilon_0 \hbar^2 \varepsilon_0}.\frac{e^2}{4\pi \varepsilon_0}\) can be replaced by \(\frac{\hbar^2}{m_e a_0 }\). This gives Eh = \(\frac{m_e e^4}{(4\pi \varepsilon_0)^2 \hbar^2}\)

10. What is the expression for energy levels of a hydrogen like atom in terms of Hartree energy?
a) E = –\(\frac{Z^2 E_h}{2n^2}\)
b) E = –\(\frac{Z^2 E_h e^2}{2n^2.4\pi \varepsilon_0}\)
c) E = –\(\frac{Z^2 E_h e^2}{2n^2}\)
d) E = –\(\frac{Z^2 E_h}{2n^2.4\pi \varepsilon_0}\)
View Answer

Answer: a
Explanation: Energy levels for a hydrogen like atom are given by E = –\(\frac{\mu e^4 Z^2}{2(4\pi \varepsilon_0)^2 \hbar^2 n^2}\). Since Hartree energy Eh = \(\frac{m_e e^4}{(4\pi \varepsilon_0)^2 \hbar^2}\), this can be substituted into the general energy equation, which results in E = –\(\frac{Z^2 E_h}{2n^2}\).
advertisement

11. What happens to the energy of a hydrogen like atom if its Hartree energy is increased by a factor of 2?
a) Stays the same
b) Increases by a factor of two
c) Decreases by a factor of two
d) Hartree energy is a physical constant with a fixed value
View Answer

Answer: d
Explanation: Hartree energy is a constant which is given by Eh = \(\frac{m_e e^4}{(4\pi \varepsilon_0)^2 \hbar^2}\) = 27.21 eV. This value is fixed and cannot be increased in any way. It is a physical constant derived from theory, and \(\frac{E_h}{2}\) is the ground state energy of a hydrogen atom.

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.