The shortest wavelength in the Balmer series | vedclass

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(D) The wavelength $\lambda$ for a transition in a hydrogen-like atom is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^

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$2$

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(D) The wavelength $\lambda$ for a transition in a hydrogen-like atom is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the shortest wavelength, the transition occurs from $n_2 = \infty$ to $n_1$.For the Balmer series of hydrogen $(Z = 1)$, $n_1 = 2$. Thus, $\frac{1}{\lambda_H} = R(1)^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$.For the Brackett series of a hydrogen-like atom with atomic number $Z$, $n_1 = 4$. Thus, $\frac{1}{\lambda_Z} = R Z^2 \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = \frac{R Z^2}{16}$.Given $\lambda_H = \lambda_Z$, we equate the expressions: $\frac{R}{4} = \frac{R Z^2}{16}$.Simplifying, $\frac{1}{4} = \frac{Z^2}{16}$, which gives $Z^2 = 4$.Therefore, $Z = 2$.

The shortest wavelength in the Balmer series of a hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is:

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