Question

In Hydrogen atom, which of the following transitions produces a spectral line of maximum frequency.

• A. $2\to 1$
• B. $6\to 2$
• C. $4\to 3$
• D. $5\to 2$

Answer 1

The correct option is A $2\to 1$

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 $6\to 2$, $\frac{1}{{\lambda }_{62}}=R[1/2^2-1/6^2]=\left(2/9\right)R=0.22R$;

For $4\to 3$, $\frac{1}{{\lambda }_{43}}=R[1/3^2-1/4^2]=0.05R$ and

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

Since frequency is inversely proportional to wavelength so $2\to 1$ transition will give the maximum frequency.