The de-Broglie wavelength ($\lambda $B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($\lambda $G) by :
$\lambda $B = 2$\lambda $G
$\lambda $B = 3$\lambda $G
$\lambda $B = $\lambda $G/2
$\lambda $B = $\lambda $G/3
Solution
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