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(B) The energy of the emitted photon is given by the Rydberg formula for the frequency of radiation: $f = c \cdot \frac{1}{\lambda} = R c \left( \frac

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$\frac{5 R c}{36}$

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(B) The energy of the emitted photon is given by the Rydberg formula for the frequency of radiation: $f = c \cdot \frac{1}{\lambda} = R c \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$ Here,the electron jumps from the third orbit $(n_i = 3)$ to the second orbit $(n_f = 2)$. Substituting the values into the formula: $f = R c \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$ $f = R c \left( \frac{1}{4} - \frac{1}{9} \right)$ $f = R c \left( \frac{9 - 4}{36} \right)$ $f = \frac{5 R c}{36}$ Therefore,the correct option is $B$.

If an electron in a hydrogen atom jumps from the third orbit to the second orbit,the frequency of the emitted radiation is given by (where $c$ is the speed of light and $R$ is the Rydberg constant):

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