Atoms & Nuclei

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60 of 239

Question 60 of 239

A hydrogen atom in its ground state absorbs 10.2eV10.2eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of (Given, Planck's constant =6.6×10−34Js=6.6\times {10}^{-34}Js).

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Answer: (B) 1.05×10−34Js1.05\times {10}^{-34}Js

Energy absorbed is given by,
ΔE=13.6×(1n12−1n22)\Delta E=13.6\times \left(\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right)
10.2=13.6×(1n12−1n22)10.2=13.6\times \left(\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right)
Therefore, we get, {n}_{1}=1&{n}_{2}=2
The change in angular momentum
ΔL=n2h2π−n1h2π=h2π\Delta L=\dfrac{{n}_{2}h}{2\pi }-\dfrac{{n}_{1}h}{2\pi }=\dfrac{h}{2\pi }
=1.05×10−34Js=1.05\times {10}^{-34}Js

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