An electron in a hydrogen-like atom has energy equal to $ -0.04 E_0 $, where $ E_0 $ is the magnitude of energy of this electron in ground state in eV. If the angular momentum of this electron is $ L $, then the value of $ \frac{2 \pi L}{h} $ is (where $ h $ is Planck's constant).

Question

The wavelength of spectral line obtained in the spectrum of Li$^{2+}$ ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

Answer 1

To solve this problem, we need to find the value of $ \frac{2 \pi L}{h} $ for an electron in a hydrogen-like atom, given that its energy is $ -0.04 E_0 $, where $ E_0 $ is the energy of the electron in the ground state.

The energy of an electron in a hydrogen-like atom is given by:

$E_n = -\frac{E_0}{n^2}$

Given, $ E_n = -0.04 E_0 $.

$-\frac{E_0}{n^2} = -0.04 E_0$

Canceling out the negative and the $ E_0 $:

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

Solving for $ n $,

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

$n = 5$

The angular momentum $ L $ of an electron in a hydrogen-like atom is quantized and given by:

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

Substituting $ n = 5 $, we get:

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

Therefore,

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

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

The correct answer is: 5.

An electron in a hydrogen-like atom has energy equal to \( -0.04 E_0 \), where \( E_0 \) is the magnitude of energy of this electron in ground state in eV. If the angular momentum of this electron is \( L \), then the value of \( \frac{2 \pi L}{h} \) is (where \( h \) is Planck's constant).

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The Correct Option is C

Solution and Explanation

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