MHT CET · Physics · Atomic Physics
If \(^{\prime} \lambda_{1}\) ' and ' \(\lambda_{2}\) ' are the wavelengths of de-Broglie waves for electrons in first and second Bohr orbits in hydrogen atom, then \(\left(\frac{\lambda_{1}}{\lambda_{2}}\right)\) is equal to (Energy in \(1^{\text {st }}\) Bohr
orbit \(=-13.6 \mathrm{eV}\) )
- A \(\frac{1}{5}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{4}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\lambda_n = \frac{2\pi r_n}{n}\) \(\lambda_n = \frac{2\pi (n^2 a_0)}{n} = 2\pi n a_0\)
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