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(D) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right

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$\frac{20}{27} \lambda$

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(D) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$,where $R$ is the Rydberg constant and $Z$ is the atomic number.For the transition from $M$-shell $(n_i = 3)$ to $L$-shell $(n_f = 2)$:$\frac{1}{\lambda} = K \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = K \left( \frac{1}{4} - \frac{1}{9} ight) = K \left( \frac{9-4}{36} ight) = \frac{5K}{36}$,where $K = R Z^2$.For the transition from $N$-shell $(n_i = 4)$ to $L$-shell $(n_f = 2)$:$\frac{1}{\lambda'} = K \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = K \left( \frac{1}{4} - \frac{1}{16} ight) = K \left( \frac{4-1}{16} ight) = \frac{3K}{16}$.Now,taking the ratio of the two equations:$\frac{\lambda'}{\lambda} = \frac{5K/36}{3K/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{5 	imes 4}{9 	imes 3} = \frac{20}{27}$.Therefore,$\lambda' = \frac{20}{27} \lambda$.

In a hydrogen-like atom,when an electron jumps from the $M$-shell to the $L$-shell,the wavelength of the emitted radiation is $\lambda$. If an electron jumps from the $N$-shell to the $L$-shell,the wavelength of the emitted radiation will be:

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