Question

In hydrogen atom which of the following transitions produces spectral line of maximum wavelength?

• A. $2\to 1$
• B. $4\to 1$
• C. $6\to 5$
• D. $5\to 2$

Answer 1

Answer: (C)

$6\to 5$

The Rydberg formula for Hydrogen atom, $\frac{1}{\lambda }=R[1/n_1^2-1/n_2^2]$ where $R=$ Rydberg constant and $n_1<n_2$

For $2\to 1$, $\frac{1}{{\lambda }_{21}}=R[1/1^2-1/2^2]=\left(3/4\right)R=0.75R$;

For $4\to 1$, $\frac{1}{{\lambda }_{41}}=R[1/1^2-1/4^2]=\left(2/9\right)R=0.94R$;

For $6\to 5$, $\frac{1}{{\lambda }_{65}}=R[1/5^2-1/6^2]=0.01R$ and

For $5\to 2$, $\frac{1}{{\lambda }_{52}}=R[1/2^2-1/5^2]=0.21R$

Thus $6\to 5$ transition will give the maximum wavelength as $1/{\lambda }_{65}$ is minimum.