The energy of an electron in an orbit of Bohr's atom is −0.04ε₀ eV, where ε₀ is the ground-state energy. If L is the angular momentum of the electron in this orbit and h is Planck's constant, then

2πL/h

is __________.

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The correct answer is 5.

Step-by-Step Explanation:

1. Energy of an electron in Bohr's orbit:
According to Bohr's model of the hydrogen atom, the energy of an electron in the n-th orbit is inversely proportional to the square of the principal quantum number n:

$E_n=\frac{E_1}{n^2}$

Here, $E_1=-{\epsilon }_0$ represents the ground-state energy of the electron. Therefore, the energy in the n-th orbit can be written as:

$E_n=-\frac{{\epsilon }_0}{n^2}$

2. Finding the principal quantum number (n):
We are given that the energy of the electron in the orbit is $-0.04{\epsilon }_0\text{eV}$. Equating the two expressions for energy:

$-\frac{{\epsilon }_0}{n^2}=-0.04{\epsilon }_0$

Dividing both sides by $-{\epsilon }_0$:

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

Taking the reciprocal on both sides gives:

$n^2=25$

Taking the positive square root (since n is a positive integer):

$n=5$

3. Applying Bohr's Angular Momentum Quantization Postulate:
According to Bohr's second postulate, the orbital angular momentum L of an electron in a stable orbit is an integral multiple of $\frac{h}{2\pi }$:

$L=\frac{nh}{2\pi }$

Rearranging this formula to express $\frac{2\pi L}{h}$:

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

Substituting $n=5$ into the equation:

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

Thus, the required value is 5.