JEE Mains · Physics · STD 12 - 12. atoms
A hydrogen atom in is ground state absorbs \(10.2\;eV\) of energy. The angular momentum of electron of the hydrogen atom will increase by the value of ............... \(\times 10^{-34} \; J {s}\) (Given, Plank's constant \(\left.=6.6 \times 10^{-34} Js \right)\)
- A \(2.10\)
- B \(1.05\)
- C \(3.15\)
- D \(4.2\)
Answer & Solution
Correct Answer
(B) \(1.05\)
Step-by-step Solution
Detailed explanation
\(13.6\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)=10.2\) \(n=2\) \(L_{i}=\frac{h}{2 \pi} \times 1\) \(L_{F}=\frac{2 h}{2 \pi}\) \(\Delta L=L_{F}-L_{i}=\frac{h}{2 \pi}=\frac{6.6 \times 10^{-34}}{2 \times \frac{22}{7}}\) \(=1.05 \times 10^{-34} \; J-s\)
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