Question

Which of the following represents the wavelength of spectral line of Balmer series of He$^+$ ion? (R = Rydberg constant, n $>$ 2)

• A. $\frac{n^2}{R(n-2)(n+2)}$
• B. $\frac{n^2}{R(n-2)(n+2)}$
  % This seems like a duplicate option - please double-check your source
• C. $\frac{n^2}{4R(n-2)(n+2)}$
• D. $\frac{n^2}{4R(n-2)(n+2)}$
  % This also seems like a duplicate - please double-check your source

Hint:

For Balmer series: $\frac{1}{\lambda} = RZ^2 \left(\frac{1}{2^2} - \frac{1}{n^2}\right)$.

Answer 1

The Correct Option is A

The wavelength of a spectral line in the hydrogen-like species (like He$^+$) is given by the Rydberg formula: $$ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$ where $R$ is the Rydberg constant, $Z$ is the atomic number, $n_1$ is the lower energy level, and $n_2$ is the higher energy level. For the Balmer series, $n_1 = 2$. For He$^+$, $Z=2$. Thus, $$ \frac{1}{\lambda} = R(2^2) \left( \frac{1}{2^2} - \frac{1}{n^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{n^2} \right) = R \left( 1 - \frac{4}{n^2} \right) = R \left( \frac{n^2 - 4}{n^2} \right) $$ Therefore, $$ \lambda = \frac{n^2}{R(n^2 - 4)} = \frac{n^2}{R(n-2)(n+2)} $$