An gas of H-atoms in excited state $n_{2}$ absorbs a photon of some energy and jump in higher energy state $n_{1}$ . Then it returns to ground state after emitting six different wavelengths in emission spectrum. The energy of emitted photon is equal. less or greater than the energy of absorbed photon then $n_{1}$ and $n_{2}$ will be

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$n_{1} = 4, n_{2} = 1$

$n_{1} = 4, n_{2} = 3$

$n_{1} = 5, n_{2} = 3$

$n_{1} = 5, n_{2} = 2$

answer is D.

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Detailed Solution

As the electron returns to ground state after emitting six different wavelengths in emission spectrum, there must be a difference of '2' between the ground and excited states. Also the ground state should be n =1. Hence $n_{1}$ = 4, $n_{2}$ =1

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