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(D) The energy difference between two orbits in a hydrogen atom is given by the Rydberg formula for wavenumber: $\frac{1}{\lambda} = R \left( \frac{1}

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$ \frac{5RC}{36} $

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(D) The energy difference between two orbits in a hydrogen atom is given by the Rydberg formula for wavenumber: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Since the frequency $f$ is related to wavelength $\lambda$ by $f = \frac{C}{\lambda}$,we can write $\frac{f}{C} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Rearranging for frequency,we get $f = RC \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.Given $n_1 = 2$ and $n_2 = 3$,we substitute these values into the equation:$f = RC \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = RC \left( \frac{1}{4} - \frac{1}{9} \right)$.Calculating the fraction: $\frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.Thus,the frequency is $f = \frac{5RC}{36}$.

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

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