The shortest wavelength of hydrogen atom in the Lyman series is. The longest wavelength in the Balmer series of He^+ is:

Question

Accepted Answer

Given: The shortest wavelength in the Lyman series of hydrogen is $\lambda$ . We need the longest wavelength in the Balmer series of He $^+$ .

Find: The corresponding expression in terms of $\lambda$ .

For hydrogen-like species, use the Rydberg relation:

\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

For the shortest wavelength in the Lyman series of hydrogen, the transition is $n_2 = \infty$ to $n_1 = 1$ with $Z = 1$ :

\frac{1}{\lambda} = R(1)^2 \left( 1 - 0 \right) = R

So,

\lambda = \frac{1}{R}

For the longest wavelength in the Balmer series of He $^+$ , the transition is $n_2 = 3$ to $n_1 = 2$ with $Z = 2$ :

\frac{1}{\lambda_{\text{Balmer}}} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)

\frac{1}{\lambda_{\text{Balmer}}} = 4R \left( \frac{1}{4} - \frac{1}{9} \right)

\frac{1}{\lambda_{\text{Balmer}}} = 4R \left( \frac{5}{36} \right) = \frac{5R}{9}

Thus,

\lambda_{\text{Balmer}} = \frac{9}{5R}

Using $\lambda = \frac{1}{R}$ ,

\lambda_{\text{Balmer}} = \frac{9}{5}\lambda

Therefore, the correct option is B. The listed the solution marks D, but its working clearly gives $\frac{9}{5}\lambda$ , which matches option B.

Answer