Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Atoms

Question:

Taking Rydberg's constant $R_H=1.097 \times 10^7~m$  first and second wavelength of Balmer series in hydrogen spectrum is

Options:

2000 Å, 3000 Å

1575 Å, 2960 Å

6529 Å, 4280 Å

6552 Å, 4863 Å

Correct Answer:

6552 Å, 4863 Å

Explanation:

$\frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$.

For first wavelength, $n_1=2$, $n_2=3 \Rightarrow \lambda_1=6563$ Å.

For second wavelength, $n_1=2, n_2=4 \Rightarrow \lambda_2=4861 ~Å$