Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Atoms

Question:

The first line of Balmer series has wavelength 6563 Å. What will be the wavelength of the first member of Lyman series :

Options:

1215.4 Å

2500 Å

7500 Å

600 Å

Correct Answer:

1215.4 Å

Explanation:

$\frac{1}{\lambda_{\text {Balmer }}}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5 R}{36}, \frac{1}{\lambda_{\text {Lyman }}}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3 R}{4}$

∴  $\lambda_{\text {Lyman }}=\lambda_{\text {Balmer }} \times \frac{5}{27}=1215.4~Å$