The angular momentum for the electron in Bohr's orbit is $$L$$. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be

In Bohr's model, the angular momentum of an electron in the $$n^{th}$$ orbit is:

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

For the first orbit: $$L_1 = \frac{h}{2\pi} = L$$ (given as L)

For the second orbit: $$L_2 = \frac{2h}{2\pi} = 2L$$

Change in angular momentum: $$\Delta L = L_2 - L_1 = 2L - L = L$$

The correct answer is Option 3: L.

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