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(B) The wavelength $\lambda$ emitted during an electronic transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} -

Vedclass · Accepted Answer

$27$

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(B) The wavelength $\lambda$ emitted during an electronic transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,where $R$ is the Rydberg constant.For the first transition: electron jumps from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$.$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$. Thus,$\lambda = \frac{36}{5R}$.For the second transition: electron jumps from the third excited state $(n_i = 4)$ to the second orbit $(n_f = 2)$.$\frac{1}{\lambda'} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right)$. Thus,$\lambda' = \frac{16}{3R}$.We are given $\lambda' = \frac{20 \lambda}{x}$. Substituting the values:$\frac{16}{3R} = \frac{20}{x} \cdot \frac{36}{5R} \implies \frac{16}{3} = \frac{20 \cdot 36}{5x} \implies \frac{16}{3} = \frac{4 \cdot 36}{x} \implies \frac{16}{3} = \frac{144}{x}$.$x = \frac{144 \cdot 3}{16} = 9 \cdot 3 = 27$.

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