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(D) The frequency $f$ of the emitted radiation is given by the Rydberg formula: $f = c \cdot \frac{1}{\lambda} = cR \left[ \frac{1}{n_1^2} - \frac{1}{

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$\frac{5RC}{36}$

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(D) The frequency $f$ of the emitted radiation is given by the Rydberg formula: $f = c \cdot \frac{1}{\lambda} = cR \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ Here,the electron jumps from $n_2 = 3$ to $n_1 = 2$. Substituting the values: $f = cR \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]$ $f = cR \left[ \frac{1}{4} - \frac{1}{9} \right]$ $f = cR \left[ \frac{9 - 4}{36} \right]$ $f = \frac{5RC}{36}$ Thus,the correct option is $D$.

If an electron in a hydrogen atom jumps from an orbit of level $n=3$ to an orbit of level $n=2$,then the emitted radiation frequency is (where $R=$ Rydberg's constant,$C=$ velocity of light).

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