What is the shortest wavelength in the Pfund | vedclass

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(D) For the $Pfund$ series,the transition occurs from $n_2 = \infty$ to $n_1 = 5$.For $He^{+}$ ion,the atomic number $Z = 2$.The Rydberg formula for w

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$25/(4R)$

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(D) For the $Pfund$ series,the transition occurs from $n_2 = \infty$ to $n_1 = 5$.For $He^{+}$ ion,the atomic number $Z = 2$.The Rydberg formula for wavelength $\lambda$ is given by: $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.Substituting the values: $\frac{1}{\lambda} = R (2)^2 (\frac{1}{5^2} - \frac{1}{\infty^2})$.$\frac{1}{\lambda} = R 	imes 4 	imes (\frac{1}{25} - 0) = \frac{4R}{25}$.Therefore,$\lambda = \frac{25}{4R}$.

What is the shortest wavelength in the $Pfund$ series of $He^{+}$ ion?

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