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NEET2026Physics-Atoms

NEET 2026 Physics Bohr Model of Hydrogen Atom MCQ Question

Type: MCQ-numerical-Medium-Class 12

In the first excited state of hydrogen atom, the energy of its electron is −3.4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately: (Take 1 eV = 1.6 × 10⁻¹⁹ J, e = 1.6 × 10⁻¹⁹ C and 14πε0=9×109 N m2/C2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2)

A

2.1 × 10⁻⁹ m

B

2.1 × 10⁻⁸ m

C

2.1 × 10⁻¹⁰ m

D

2.1 × 10⁻¹¹ m

Correct Answer

Option C

Detailed Explanation

To determine the radial distance of the electron from the hydrogen nucleus in the first excited state of the hydrogen atom, we can use the concepts from the Bohr model of the hydrogen atom. According to the Bohr model, the energy levels of the hydrogen atom are given by the formula:

En=−13.6 eVn2E_n = -\frac{13.6 \, \text{eV}}{n^2}

where EnE_n is the energy of the electron in the nth energy level, and nn is the principal quantum number. For the first excited state, n=2n = 2, so we can calculate the energy:

E2=−13.6 eV22=−13.6 eV4=−3.4 eVE_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV}

This confirms that the energy of the electron in the first excited state is indeed −3.4 eV-3.4 \, \text{eV}.

Step 1: Convert Energy to Joules

We need to convert the energy from electron volts to joules for further calculations. Using the conversion 1 eV=1.6×10−19 J1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}:

E2=−3.4 eV×1.6×10−19 J/eV=−5.44×10−19 JE_2 = -3.4 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = -5.44 \times 10^{-19} \, \text{J}

Step 2: Calculate the Radial Distance

Using the Bohr model, the radius of the electron's orbit for any energy level nn can be expressed as:

rn=n2a0r_n = n^2 a_0

where a0a_0 (the Bohr radius) is approximately 5.29×10−11 m5.29 \times 10^{-11} \, \text{m}. For the first excited state (n=2n = 2):

r2=22⋅a0=4⋅(5.29×10−11 m)=2.116×10−10 mr_2 = 2^2 \cdot a_0 = 4 \cdot (5.29 \times 10^{-11} \, \text{m}) = 2.116 \times 10^{-10} \, \text{m}

Step 3: Round and Compare

Rounding this to two significant figures gives us:

r2≈2.1×10−10 mr_2 \approx 2.1 \times 10^{-10} \, \text{m}

Conclusion

Now, we check the options provided:

  • A) 2.1×10−9 m2.1 \times 10^{-9} \, \text{m} — Incorrect, too large.
  • B) 2.1×10−8 m2.1 \times 10^{-8} \, \text{m} — Incorrect, too large.
  • C) 2.1×10−10 m2.1 \times 10^{-10} \, \text{m} — Correct, matches our calculation.
  • D) 2.1×10−11 m2.1 \times 10^{-11} \, \text{m} — Incorrect, too small.

Thus, the correct answer is C) 2.1×10−10 m2.1 \times 10^{-10} \, \text{m}.

Summary

The radial distance of the electron from the hydrogen nucleus in the first excited state is approximately 2.1×10−10 m2.1 \times 10^{-10} \, \text{m}. The calculation involved converting energy from eV to Joules, applying the Bohr model radius formula, and confirming that the only valid option matching our result is option C.

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