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(D) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.For an electron in the $n^{th}$

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will increase.

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(D) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.For an electron in the $n^{th}$ orbit of a hydrogen atom,the velocity $v$ is inversely proportional to the principal quantum number $n$ $(v \propto \frac{1}{n})$.Since momentum $p = mv$,we have $p \propto \frac{1}{n}$.Substituting this into the de-Broglie relation,we get $\lambda = \frac{h}{p} \propto n$.The ground state corresponds to $n = 1$. The third excited state corresponds to $n = 4$.Since the principal quantum number $n$ increases from $1$ to $4$,the de-Broglie wavelength $\lambda$ will increase.

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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