The wavelength of the radiation emitted is la | vedclass

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

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the first case: The electron jumps from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$.$\frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight)$.So, $\lambda_0 = \frac{36}{5R}$.For the second case: The electron jumps from the third excited state $(n_i = 4)$ to the second orbit $(n_f = 2)$.$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = R \left( \frac{3}{16} ight)$.So, $\lambda = \frac{16}{3R}$.Now, express $\lambda$ in terms of $\lambda_0$:$\lambda = \frac{16}{3R} = \frac{16}{3} 	imes \frac{5\lambda_0}{36} = \frac{80\lambda_0}{108} = \frac{20\lambda_0}{27}$.Comparing this with the given expression $\frac{20\lambda_0}{x}$, we get $x = 27$.

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