If hydrogen atom is its ground state absorbs, 10.2 eV of energy. The orbital angularmomentum is increase by

Correct answer: (A) 1.05 × 10^-34 ~J / s. Electron goes to its first excited state (n=2) from ground state (n=1) after absorbing 10.2 eV energy ∴ Increase in momentum =h/(2 π) =frac6.6 × 10^-346.28 =1.05 × 10^-34 ~J-S

Q106 / 120 · VITEEE 2012 · Physics

If hydrogen atom is its ground state absorbs, $10.2 \mathrm{eV}$ of energy. The orbital angularmomentum is increase by

Correct answer
A
$1.05 \times 10^{-34} \mathrm{~J} / \mathrm{s}$
B
$3.16 \times 10^{-34} \mathrm{~J} / \mathrm{s}$
C
$2.11 \times 10^{-34} \mathrm{~J} / \mathrm{s}$
D
$4.22 \times 10^{-34} \mathrm{~J} / \mathrm{s}$

Solution

Electron goes to its first excited state $(\mathrm{n}=2)$ from ground state $(\mathrm{n}=1)$ after absorbing $10.2 \mathrm{eV}$ energy $$ \begin{array}{l} \therefore \text { Increase in momentum }=\frac{\mathrm{h}}{2 \pi} \\ =\frac{6.6 \times 10^{-34}}{6.28} \\ =1.05 \times 10^{-34} \mathrm{~J}-\mathrm{S} \end{array} $$

Hence, the correct answer is option: 'A'.

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