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

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 m_{e} and mass of the nucleus is m_{n}, 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

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

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.

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

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

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 a_{0} = \(\frac{h^2 \varepsilon_0}{\pi m_e e^2}\) in picometers (pm)?

a) 203.32 pm

b) 12.45 pm

c) 52.92 pm

d) 103.84 pm

View Answer

Explanation: Evaluating all the constants in a

_{0}= \(\frac{h^2 \varepsilon_0}{\pi m_e e^2}\) in terms of SI units gives Bohr’s radius as = 52.9177 × 10

^{-12}m. 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 E_{h} = \(\frac{e^2}{4\pi \varepsilon_0 a_0}\) in eV?

a) 27.21 eV

b) 103.43 eV

c) 90.99 eV

d) 132.24 eV

View Answer

Explanation: Plugging in the value of all physical constants in the expression E

_{h}= \(\frac{e^2}{4\pi \varepsilon_0 a_0}\) gives E

_{h}= 4.360 × 10

^{-18}J = 27.21 eV. This is a very important constant because the energy of a hydrogen atom in its ground state is equal to \(\frac{E_h}{2}\).

9. The expression for Hartree energy is E_{h} = \(\frac{e^2}{4\pi \varepsilon_0 a_0}\). What is its expression purely in terms of fundamental constants, eliminating Bohr’s radius?

a) E_{h} = \(\frac{m_e e}{(4\pi \varepsilon_0)^2 \hbar}\)

b) E_{h} = \(\frac{m_e e^4}{\hbar^2}\)

c) E_{h} = \(\frac{m_e e^4 \hbar^2}{(4\pi \varepsilon_0)^2}\)

d) E_{h} = \(\frac{m_e e^4}{(4\pi \varepsilon_0)^2 \hbar^2}\)

View Answer

Explanation: Generally, E

_{h}= \(\frac{e^2}{4\pi \varepsilon_0 a_0}\) and a

_{0}= \(\frac{h^2 \varepsilon_0}{\pi m_e e^2}\). Plugging the value of Bohr’s radius in Hartree energy gives E

_{h}= \(\frac{e^2 \pi m_e e^2}{4\pi \varepsilon_0 \hbar^2 \varepsilon_0}.\frac{e^2}{4\pi \varepsilon_0}\) can be replaced by \(\frac{\hbar^2}{m_e a_0 }\). This gives E

_{h}= \(\frac{m_e e^4}{(4\pi \varepsilon_0)^2 \hbar^2}\)

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

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 E

_{h}= \(\frac{m_e e^4}{(4\pi \varepsilon_0)^2 \hbar^2}\), this can be substituted into the general energy equation, which results in E = –\(\frac{Z^2 E_h}{2n^2}\).

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

Explanation: Hartree energy is a constant which is given by E

_{h}= \(\frac{m_e e^4}{(4\pi \varepsilon_0)^2 \hbar^2}\) = 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.

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