The de-Broglie wavelength associated with electron in the $n = 4$ level is

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two times the de-Broglie wavelength of the electron in the ground state

four times the de-Broglie wavelength of the electron in the ground state

half of the de-Broglie wavelength of the electron in the ground state

$\frac{1}{4}$ th of the de-Broglie wavelength of the electron in the ground state.

answer is B.

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Detailed Solution

de-Broglie wavelength of electron, $λ = \frac{h}{m v} \dots (i)$

Also angular momentum is quantised, i.e., $m v r = \frac{n h}{2 π} \dots (ii)$

Using (i) and (ii), $λ = \frac{2 π r}{n}$

Since $r \propto n^{2}$ , $λ \propto n$ .

For $n = 4, λ_{4} = 4 λ_{1}$ , i.e., the de-Broglie wavelength is four times that of ground state.

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