JEE PYQ: Atomic Physics Question 61
Question 61 - 2019 (11 Jan Shift 2)
In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:
(1) $\frac{27}{20}\lambda$
(2) $\frac{16}{25}\lambda$
(3) $\frac{25}{16}\lambda$
(4) $\frac{20}{27}\lambda$
Show Answer
Answer: (4)
Solution
When electron jumps from M $\to$ L shell: $\frac{1}{\lambda} = K\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = \frac{K \times 5}{36}$ …(i). When electron jumps from N $\to$ L shell: $\frac{1}{\lambda’} = K\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = \frac{K \times 3}{16}$ …(ii). Solving equations (i) and (ii): $\lambda’ = \frac{20}{27}\lambda$.
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