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Question

(A) The frequency of a photon emitted during a transition is given by Rydberg's formula: $f = R c Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

Vedclass · Accepted Answer

$\frac{243}{32}f$

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(A) The frequency of a photon emitted during a transition is given by Rydberg's formula: $f = R c Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For a hydrogen atom $(Z=1)$,the transition from the $2^{nd}$ excited state $(n_2=3)$ to the ground state $(n_1=1)$ gives frequency $f$:$f = R c (1)^2 \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = R c \left( 1 - \frac{1}{9} ight) = \frac{8}{9} R c$.Thus,$R c = \frac{9}{8} f$.For $Li^{++}$ $(Z=3)$,the transition from the $1^{st}$ excited state $(n_2=2)$ to the ground state $(n_1=1)$ gives frequency $f'$:$f' = R c (3)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R c (9) \left( 1 - \frac{1}{4} ight) = R c (9) \left( \frac{3}{4} ight) = \frac{27}{4} R c$.Substituting $R c = \frac{9}{8} f$ into the equation for $f'$:$f' = \frac{27}{4} 	imes \left( \frac{9}{8} f ight) = \frac{243}{32} f$.

When an electron in a hydrogen atom makes a transition from the $2^{nd}$ excited state to the ground state,it emits a photon of frequency $f$. The frequency of the photon emitted when an electron of $Li^{++}$ makes a transition from the $1^{st}$ excited state to the ground state is:

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