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

$H{e}^{+}$ion :
$\frac{1}{\lambda (H)}=R(1{)}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$\frac{1}{\lambda \left(H{e}^{+}\right)}=R(2{)}^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
Given $\lambda (H)=\lambda \left(H{e}^{+}\right)$
$R(1{)}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]=R(4)\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
$\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{1^2}-\frac{1}{2^2}$
On comparing $n_1=1\&n_2=2$

Answer 2

The wavelength for the transitions in hydrogen is given by the Rydberg formula:

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

where \( R(1) \) is the Rydberg constant for hydrogen. For He\(^{+}\), the formula is:

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

Given \( \lambda(H) = \lambda(\text{He}^+) \), we equate the two equations:

\[ R(1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Simplifying and comparing \( n_1 = 1 \) and \( n_2 = 2 \), we find the correct transition in the hydrogen spectrum is from \( n = 2 \) to \( n = 1 \). Thus, the correct answer is option (1).