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MCQEasyJEE 2023Bohr's Model & Hydrogen Spectrum

JEE Physics 2023 Question with Solution

Speed of an electron in Bohr's 7th7^{\text{th}} orbit for Hydrogen atom is 3.6×106 m/s3.6 \times 10^6 \, \text{m/s}. The corresponding speed of the electron in the 3rd3^{\text{rd}} orbit, in m/s\text{m/s}, is:

  • A

    1.8×106 m/s1.8 \times 10^6 \, \text{m/s}

  • B

    7.5×106 m/s7.5 \times 10^6 \, \text{m/s}

  • C

    3.6×106 m/s3.6 \times 10^6 \, \text{m/s}

  • D

    8.4×106 m/s8.4 \times 10^6 \, \text{m/s}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The speed of the electron in the 7th7^{\text{th}} orbit is 3.6×106 m/s3.6 \times 10^6 \, \text{m/s}.

Find: The speed of the electron in the 3rd3^{\text{rd}} orbit.

For a hydrogen atom, the electron speed varies inversely with the principal quantum number nn.

v1v2=n2n1\frac{v_1}{v_2} = \frac{n_2}{n_1}

Using v7v_7 and v3v_3:

v7v3=37\frac{v_7}{v_3} = \frac{3}{7}

Substituting the given value:

3.6×106v3=37\frac{3.6 \times 10^6}{v_3} = \frac{3}{7}

Solving for v3v_3:

v3=3.6×106×73=8.4×106 m/sv_3 = \frac{3.6 \times 10^6 \times 7}{3} = 8.4 \times 10^6 \, \text{m/s}

Therefore, the speed of the electron in the 3rd3^{\text{rd}} orbit is 8.4×106 m/s8.4 \times 10^6 \, \text{m/s}. The correct option is D.

The solution labels the correct option as B, but the worked value clearly matches option D.

Direct Ratio Shortcut

Given: Electron speed in the 7th7^{\text{th}} orbit is 3.6×106 m/s3.6 \times 10^6 \, \text{m/s}.

Find: Electron speed in the 3rd3^{\text{rd}} orbit.

Since v∝1nv \propto \frac{1}{n}, speed increases when nn decreases. So,

v3=v7×73v_3 = v_7 \times \frac{7}{3}

Substitute the value:

v3=3.6×106×73=8.4×106 m/sv_3 = 3.6 \times 10^6 \times \frac{7}{3} = 8.4 \times 10^6 \, \text{m/s}

Therefore, the correct option is D.

Common mistakes

  • Using the proportionality in the wrong direction. Since v∝1nv \propto \frac{1}{n}, a smaller orbit number means a larger speed. Do not write v∝nv \propto n; instead use the inverse relation carefully.

  • Matching the numerical result to the wrong option because of the inconsistent label in the solution. Always compare the final computed value 8.4×106 m/s8.4 \times 10^6 \, \text{m/s} with the listed options and then choose the corresponding option.

  • Substituting n1n_1 and n2n_2 incorrectly in the ratio formula. Keep orbit and speed pairs aligned: v7v_7 must go with n=7n=7 and v3v_3 must go with n=3n=3.

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