Question:
The de-Broglie wavelength (\(\lambda_B\)) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (\(\lambda_G\)) by :
The de-Broglie wavelength (\(\lambda_B\)) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (\(\lambda_G\)) by :
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The second excited state is \(n=3\); \(\lambda\propto n\) in the Bohr model.
Updated On: Nov 11, 2026
- \(\lambda_B=2\lambda_G\)
- \(\lambda_B=3\lambda_G\)
- \(\lambda_B=\lambda_{G/2}\)
- \(\lambda_B=\lambda_{G/3}\)
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The Correct Option is B
Solution and Explanation
Bohr's quantisation: \(mvr=\frac{nh}{2\pi}\Rightarrow \lambda=\frac{h}{mv}=\frac{2\pi r}{n}\).
With \(r_n=n^2r_1\): \(\lambda_n=\frac{2\pi n^2r_1}{n}=n\lambda_1\).
Second excited state: \(n=3\), so \(\lambda_B=3\lambda_G\).
Answer: (2) \(\lambda_B=3\lambda_G\)
With \(r_n=n^2r_1\): \(\lambda_n=\frac{2\pi n^2r_1}{n}=n\lambda_1\).
Second excited state: \(n=3\), so \(\lambda_B=3\lambda_G\).
Answer: (2) \(\lambda_B=3\lambda_G\)
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