Atoms & Nuclei

JEE Main level, numerical answer, medium

53 of 239

Question 53 of 239

A light of energy 12.75eV12.75eV 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 xπ×10−17eVs\dfrac{x}{\pi }\times {10}^{-17}eVs. The value of xx is ______ (use h=4.14×10–15eVs,c=3×108ms–1h=4.14\times {10}^{–15}eVs,c=3\times {10}^{8}m{s}^{–1})

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Answer: 828

Let the electron jumped to nth{n}^{th} excited state.
In the ground state, energy E=−13.6eVE=-13.6eV
So, using relation 12.75=13.6[112−1n2]12.75=13.6\left[\dfrac{1}{{1}^{2}}-\dfrac{1}{{n}^{2}}\right]
⇒0.9375=[1−1n2]⇒n2=16\Rightarrow 0.9375=\left[1-\dfrac{1}{{n}^{2}}\right] \Rightarrow {n}^{2}=16
⇒n=16=4\Rightarrow n=\sqrt{16}=4
Now, angular momentum, L=nh2π=4h2π=2hπL=\dfrac{nh}{2\pi }=\dfrac{4h}{2\pi }=\dfrac{2h}{\pi }
=2π×4.14×10−15=\dfrac{2}{\pi }\times 4.14\times {10}^{-15}
=828×10−17πeVs=\dfrac{828\times {10}^{-17}}{\pi }eVs
Hence, the value of x=828x=828.

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