Question

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $n =4$ to $n =2$ of $He ^{+}$spectrum

• A. $n =1$ to $n =3$
• B. $n =1$ to $n =2$
• C. $n =2$ to $n =1$
• D. $n =3$ to $n =4$

Hint:

The wavelengths of transitions in the hydrogen and He\(^{+}\) spectra can be compared using the Rydberg formula, considering the differences in the ionization energies of the atoms.

Answer 1

The Correct Option is C

The problem involves determining the transition in the hydrogen spectrum that would have the same wavelength as the Balmer type transition from \( n = 4 \) to \( n = 2 \) of the \( \text{He}^{+} \) spectrum.

To solve this, we can use the formula for the wavelength of emitted or absorbed radiation in hydrogen-like atoms:

\[\frac{1}{\lambda} = R \left(Z^2\right) \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]

where:

• \( \lambda \) is the wavelength.
• \( R \) is the Rydberg constant (\(1.097 \times 10^7\) m^-1).
• \( Z \) is the atomic number ( \( Z = 2 \) for \( \text{He}^{+} \) and \( Z = 1 \) for hydrogen).
• \( n_1 \) and \( n_2 \) are the principal quantum numbers with \( n_2 > n_1 \).

First, calculate the wavelength for the \( \text{He}^{+} \) transition from \( n = 4 \) to \( n = 2 \):

\[\frac{1}{\lambda} = R(2^2) \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4R \left(\frac{1}{4} - \frac{1}{16}\right)\]

 

\[\Rightarrow \frac{1}{\lambda} = 4R \times \frac{3}{16} = \frac{3R}{4}\]

Now, for hydrogen, evaluate each given transition and see which gives the same wavelength.

• \( n = 2 \) to \( n = 1 \) (Option): 
   

\[\frac{1}{\lambda} = R(1^2) \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(1 - \frac{1}{4}\right) = R \times \frac{3}{4} = \frac{3R}{4}\]

•  
  This is the same as the calculated wavelength for the \( \text{He}^{+} \) spectrum.
• \( n = 1 \) to \( n = 3 \), \( n = 1 \) to \( n = 2 \), and \( n = 3 \) to \( n = 4 \): 
   

\[\text{For these transitions, } \frac{1}{\lambda} \neq \frac{3R}{4}.\]

•  
  Hence, they do not match the He transition.

Based on this analysis, the correct answer is the transition \( n = 2 \) to \( n = 1 \) in the hydrogen spectrum, which matches the wavelength of the \( \text{He}^{+} \) transition from \( n = 4 \) to \( n = 2 \).