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(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.Since $p = mv

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$1/4$

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(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.Since $p = mv$ and kinetic energy $E_K = \frac{p^2}{2m}$,we have $p = \sqrt{2mE_K}$.In a hydrogen atom,the kinetic energy of an electron in the $n^{th}$ orbit is proportional to $\frac{1}{n^2}$,i.e.,$E_K \propto \frac{1}{n^2}$.Therefore,the momentum $p = \sqrt{2mE_K} \propto \sqrt{\frac{1}{n^2}} = \frac{1}{n}$.Since $\lambda = \frac{h}{p}$,it follows that $\lambda \propto n$.When the electron jumps from the third excited state $(n = 4)$ to the ground state $(n = 1)$,the ratio of the new wavelength $\lambda_{final}$ to the initial wavelength $\lambda_{initial}$ is $\frac{\lambda_{final}}{\lambda_{initial}} = \frac{n_{final}}{n_{initial}} = \frac{1}{4}$.Thus,the de-Broglie wavelength becomes $\frac{1}{4}$ of its initial value.

When an electron in a hydrogen atom jumps from the third excited state to the ground state,the de-Broglie wavelength associated with the electron becomes

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