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(D) The velocity of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $v \propto \frac{1}{n}$.Since the momentum $p = mv$, we have $p \

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will increase.

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(D) The velocity of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $v \propto \frac{1}{n}$.Since the momentum $p = mv$, we have $p \propto \frac{1}{n}$.The de-Broglie wavelength is given by the relation $\lambda = \frac{h}{p}$.Substituting the proportionality for momentum, we get $\lambda \propto n$.As the electron jumps to a higher excited state, the principal quantum number $n$ increases.Therefore, the de-Broglie wavelength $\lambda$ will increase.

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

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