A light of energy 12.75 eV is incident on a hydrogen atom in its… — Solution | JEE Physics

NVAEasyJEE 2023Bohr's Model & Hydrogen Spectrum

A light of energy $12.75 \, \text{eV}$ 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 $\frac{x}{\pi} \times 10^{-17}$ eVs. The value of $x$ is _____.

Answer

Correct answer:828

Step-by-step solution

Standard Method

Given: Energy of incident light is $12.75 \, \text{eV}$ . The hydrogen atom is initially in the ground state.

Find: The value of $x$ in the angular momentum expression.

For hydrogen atom, energy in ground state is

E_1 = -13.6 \, \text{eV}

After absorbing the radiation,

-\frac{13.6}{n^2} = -13.6 + 12.75

So,

-\frac{13.6}{n^2} = -0.85

Hence,

n = \sqrt{16}

n = 4

Now angular momentum in the excited state is

L = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}

Using

h = 4.14 \times 10^{-15} \, \text{eVs}

we get

L = \frac{2}{\pi} \times 4.14 \times 10^{-15}

L = \frac{828 \times 10^{-17}}{\pi} \, \text{eVs}

Therefore, the value of $x$ is $828$ .

Common mistakes

Using the absorbed energy directly as the final state energy is incorrect. The atom starts from $-13.6 \, \text{eV}$ , so the photon energy must be added to the initial energy level, not treated as the new level itself.
Forgetting that Bohr angular momentum is $L = \frac{nh}{2\pi}$ is a common mistake. Using $nh$ or $\frac{h}{2\pi}$ without the factor $n$ gives the wrong value.
Taking $n = 16$ instead of $n = 4$ is wrong. From $n^2 = 16$ , the correct principal quantum number is $n = 4$ .

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