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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 = m

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$\frac{1}{4}$

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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 $p = \sqrt{2mE_K}$. Thus,$\lambda = \frac{h}{\sqrt{2mE_K}}$. For an electron in a hydrogen atom,the velocity $v$ in the $n^{	ext{th}}$ orbit is given by $v_n = \frac{v_0 Z}{n}$. Since $p = mv$,the momentum $p_n$ is proportional to $\frac{1}{n}$. Therefore,$\lambda_n = \frac{h}{p_n} \propto n$. When the electron jumps from the third excited state $(n = 4)$ to the ground state $(n = 1)$,the ratio of the wavelengths is $\frac{\lambda_{n=1}}{\lambda_{n=4}} = \frac{1}{4}$.

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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