AIIMS2018Physics-Atomic Structure

AIIMS 2018 Physics Hydrogen Spectrum MCQ Question

Type: MCQ-numerical-Medium-Class 11

If radiation corresponding to first line of “Balmer series” of He+\text{He}^+ ion knocked out electron from 1st1^{\text{st}} excited state of H\text{H} atom, the kinetic energy of ejected electron from H\text{H} atom would be (eV\text{eV})

[Given En=Z2n2(13.6 eV)]\left[\text{Given } E_n = \frac{Z^2}{n^2}(13.6\text{ eV})\right]

A

4.155 eV4.155\text{ eV}

B

8.310 eV8.310\text{ eV}

C

2.515 eV2.515\text{ eV}

D

5.550 eV5.550\text{ eV}

Correct Answer

Option A

Detailed Explanation

Option A is correct because the calculation accurately determines the energy required to eject an electron from the n = 2 level of a hydrogen atom (H) using the formula for energy levels in the hydrogen atom, En=13.6eVn2E_n = -\frac{13.6 \, \text{eV}}{n^2}. The energy difference between the n = 2 and n = ∞ states is calculated as E=13.6(140)=13.6eVE = 13.6 \left( \frac{1}{4} - 0 \right) = 13.6 \, \text{eV}, while the energy required to eject the electron from n = 2 is E=13.6(1419)=4.155eVE = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 4.155 \, \text{eV}. Other options are not provided, but any incorrect options would likely misrepresent the energy calculations or use incorrect values, leading to erroneous results. Understanding these energy transitions is crucial in quantum mechanics and atomic physics, particularly in the context of electron configurations and spectral lines.

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