A light of energy 12.75 eV is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is . The value of is _________ (use )

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Solution

Q.
A light of energy 12.75 eV is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{π} \times 10^{- 17} eV s$ . The value of $x$ is _________

(use $h = 4.14 \times 10^{- 15} eVs, c = 3 \times 10^{8} {ms}^{- 1}$ ) [2023]

Ans.
(828)

$In the ground state energy = - 13.6 eV$

$So energy \frac{- 13.6 eV}{n^{2}} = - 13.6 + 12.75$

$\frac{- 13.6 eV}{n^{2}} = - 0.85$

$n = \sqrt{16} = 4$

$Angular momentum = \frac{n h}{2 π} = \frac{4 h}{2 π} = \frac{2 h}{π}$

$Angular momentum = \frac{2}{π} \times 4.14 \times 10^{- 15}$ $= \frac{828 \times 10^{- 17}}{π} eVs$

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