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(D) The frequency of emitted radiation in a hydrogen-like ion is given by the Rydberg formula: $f = R c Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \

Vedclass · Accepted Answer

$2.46$

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(D) The frequency of emitted radiation in a hydrogen-like ion is given by the Rydberg formula: $f = R c Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$,where $k = R c Z^2$ is a constant for a specific ion.For the transition from $n=3$ to $n=1$: $f_1 = k \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = k \left( 1 - \frac{1}{9} ight) = k \left( \frac{8}{9} ight) = 2.92 	imes 10^{15} 	ext{ Hz}$.For the transition from $n=2$ to $n=1$: $f_2 = k \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = k \left( 1 - \frac{1}{4} ight) = k \left( \frac{3}{4} ight)$.Taking the ratio: $\frac{f_1}{f_2} = \frac{k(8/9)}{k(3/4)} = \frac{8}{9} 	imes \frac{4}{3} = \frac{32}{27}$.Therefore,$f_2 = f_1 	imes \frac{27}{32} = 2.92 	imes 10^{15} 	imes \frac{27}{32} \approx 2.46 	imes 10^{15} 	ext{ Hz}$.

$A$ particular hydrogen-like ion emits radiation of frequency $2.92 \times 10^{15} \text{ Hz}$ when it makes a transition from $n=3$ to $n=1$. The frequency in $\text{Hz}$ of radiation emitted in the transition from $n=2$ to $n=1$ will be: (in $\times 10^{15}$)

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