Question

The energy of an electron in an orbit of the Bohr's atom is $-0.04 E_g$ where $E_g$ 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. 4
• B. 6
• C. 5
• D. 2

Hint:

The term $\frac{2 \pi L}{h}$ is simply equal to the principal quantum number $n$ of the orbit in the Bohr model.

Answer 1

The Correct Option is C

To solve this problem, we need to understand the relationship between the energy of an electron, its angular momentum, and other constants in the Bohr model of the atom.

The energy of an electron in an orbit of Bohr's atom is given as -0.04 E_g, where E_g is the ground state energy. In the Bohr model, the energy of an electron at any level n is given by:

E_n = -\frac{E_g}{n^2}

Matching this with the given energy:

-\frac{E_g}{n^2} = -0.04 E_g

Cancelling the negative signs and E_g from both sides:

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

This gives us:

n^2 = \frac{1}{0.04} = 25

Thus, n = \sqrt{25} = 5.

The Bohr model defines the angular momentum L of an electron as:

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

Substituting the value of n = 5:

L = 5\frac{h}{2\pi}

We are interested in \frac{2\pi L}{h}:

\frac{2\pi L}{h} = \frac{2\pi \cdot 5\frac{h}{2\pi}}{h}

Simplifying this expression:

\frac{2\pi \cdot 5h}{2\pi \cdot h} = 5

Therefore, the value of \frac{2\pi L}{h} is 5.

Thus, the correct answer is:

5