Question

A hydrogen atom in the ground state is excited by monochromatic radiation of wavelength $\lambda \dot{A}$. The resulting spectrum consists of maximum 15 different lines. What is the wavelength $\lambda$?

• A. $450.63\dot{A}$
• B. $635.38\dot{A}$
• C. $937.3\dot{A}$
• D. None of these

Answer 1

The correct option is C $937.3\dot{A}$

Since the number of different lines is 15, the excited state must be n = 6.

As only then the number of lines in Lyman = 5, Balmer = 4, Paschen = 3, Brackett = 2, Pfund = 1.

Sum of those different lines = 5 + 4 + 3 + 2 + 1 = 15.

So frequency,

$\nu =3.29\times {10}^{15}(\frac{1}{1^2}-\frac{1}{6^2})Hz=3.20\times {10}^{15}Hz$

Now, wavelength $\left(\lambda \right)=\frac{c}{\nu }=\frac{3\times {10}^8}{3.20\times {10}^{15}}=9.37\times {10}^{-8}m=937.3A^0$