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(B) The de-Broglie wavelength $\lambda$ associated with an electron in the $n^{th}$ orbit of a hydrogen atom is given by the relation $\lambda = \frac

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will increase

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(B) The de-Broglie wavelength $\lambda$ associated with an electron in the $n^{th}$ orbit of a hydrogen atom is given by the relation $\lambda = \frac{h}{p} = \frac{2\pi r_n}{n}$, where $r_n$ is the radius of the $n^{th}$ orbit.Since $r_n \propto n^2$, we have $\lambda \propto \frac{n^2}{n} = n$.In the ground state, $n_1 = 1$.The third excited state corresponds to $n_2 = 4$.Since $n_2 > n_1$, the de-Broglie wavelength $\lambda$ increases as the electron moves to a higher energy state.

When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it

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