CUET · PHYSICS · PYQ PAPER 2023
When the electron in a hydrogen atom jumps from the 2nd orbit to the 1st orbit, the wavelength of the radiation emitted is \(\lambda\).
When the electron jumps from the 3rd orbit to the 1st orbit, the wavelength of emitted radiation is:
- A \(\frac{27}{32} \lambda\)
- B \(\frac{27}{5} \lambda\)
- C \(\frac{4}{5} \lambda\)
- D \(\frac{2}{3} \lambda\)
Answer & Solution
Correct Answer
(A) \(\frac{27}{32} \lambda\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(\frac{3}{4}\right)\) \(\frac{1}{\lambda'} = R \left(\frac{1}{1^2} - \frac{1}{3^2}\right) = R \left(\frac{8}{9}\right)\)…
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