JEE Main 2022 · 27 Jun · Shift 1 · Physics Q48

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PhysicsSingle correct+4 / -1Question 48 of 90

A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of :

(Given, Planck's constant $=6.6\times10^{-34}$ Js).

2.10\times10^{-34}

1.05\times10^{-34}

3.15\times10^{-34}

4.2\times10^{-34}

Answer and worked solution

Official answer: B

Transition. 10.2 eV takes the electron from $n=1$ to $n=2$ ; $L$ rises by $\dfrac h{2\pi}=\dfrac{6.6\times10^{-34}}{2\pi}\approx1.05\times10^{-34}$ J s.

(B)

Topics:Bohr model and hydrogen spectrum

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JEE Main 2022 · 27 Jun · Shift 1, Physics question 48