Question

The longest wavelength in A^o in the Balmer series of singly ionised helium ion, He⁺ is

• (A) 6565 A^o
• (B) 912 A^o
• (C) 60932.2 A^o
• (D) 1641 A^o

Answer 1

The Correct Option is D

Explanation:
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]z^2\text{ [For Balmer series }{n}_1=2,\text{ For minimum energy }{n}_2=3\}$$=10678[\frac{1}{22}-\frac{1}{32}]2^2=10678[\frac{1}{4}-\frac{1}{9}]\times 4$$=10678\left[\frac{9-4}{36}\right]=10678\times \frac{5}{9}$$\lambda =1641.1\times {10}^{-8}\mathrm{cm}=1641.1{\mathrm{A}}^{\circ }$Hence, it is the correct answer.

Answer 2

The spectral lines emitted due to the transition of an electron from the higher energy state to a particular lower energy state form a spectral series.

Various spectral series are

• Lyman Series: The spectral lines emitted due to the transition of an electron from any higher orbit to the first orbit form a spectral series known as the Lyman series.
• Balmer Series: The spectral lines emitted due to the transition of an electron from any higher orbit to the second orbit form a spectral series known as the Balmer series.
• Paschen Series: The spectral lines emitted due to the transition of an electron from any higher orbit to the third orbit form a spectral series known as the Paschen series.
• Brackett Series: The spectral lines emitted due to the transition of an electron from any higher orbit to the fourth orbit form a spectral series known as the Brackett series.
• Pfund Series: The spectral lines emitted due to the transition of an electron from any higher orbit to the fifth orbit form a spectral series known as the Pfund series.

The longest wavelength and shortest wavelength of these spectral series can be calculated by using the Rydberg formula