The energy of an electron in an orbit of the Bohr's atom is −0.04E₀ eV where E₀ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then 2πL/h is ______ :
Q. The energy of an electron in an orbit of the Bohr's atom is −0.04E₀ eV where E₀ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then \( \frac{2\pi L}{h} \) is ______ :
A. 6
B. 2
C. 5
D. 4
Correct Answer: 5

Explanation (Complete Step-by-Step Derivation)

In Bohr's model, energy of nth orbit is given by:

\[ E_n = \frac{E_1}{n^2} \]

Where ground state energy:

\[ E_1 = -E_0 \]

Given energy:

\[ E_n = -0.04E_0 \]

Substitute in formula:

\[ \frac{-E_0}{n^2} = -0.04E_0 \]

Cancel −E₀ from both sides:

\[ \frac{1}{n^2} = 0.04 \]

\[ n^2 = \frac{1}{0.04} \]

\[ n^2 = 25 \]

\[ n = 5 \]

In Bohr's model, angular momentum is quantized:

\[ L = n\frac{h}{2\pi} \]

Now calculate:

\[ \frac{2\pi L}{h} \]

Substitute L:

\[ \frac{2\pi}{h} \times n\frac{h}{2\pi} \]

\[ = n \]

\[ = 5 \]

Final Answer: 5

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