The de-Broglie wavelength $(λ_{E})$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $(λ_{G})$ by

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$λ_{E} = \frac{λ_{G}}{3}$

$λ_{E} = 3 λ_{G}$

$λ_{E} = \frac{λ_{G}}{2}$

$λ_{E} = 2 λ_{G}$

answer is B.

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Detailed Solution

For ground state, n_G=1 and for second excited state, n_E=3

As de-Broglie wavelength is given by

$λ = \frac{h}{p} \Rightarrow \frac{λ_{E}}{λ_{G}} = \frac{p_{G}}{p_{E}} = \frac{m v_{G}}{m v_{E}}$

Since, $v \propto \frac{Z}{n}$

$\Rightarrow \frac{λ_{E}}{λ_{G}} = \frac{n_{E}}{n_{G}} = \frac{3}{1} \Rightarrow λ_{E} = 3 λ_{G}$

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