Question

In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the bracket series is :

• A. 16λ
• B. 2λ
• C. 4λ
• D. 9λ

Answer 1

Answer: (C)

4λ

Answer 2

The shortest wavelength in the Balmer series occurs when an electron transitions from $\infin\ \text{to}\ n = 2.$
$\because\frac{1}{λ}=Rz^2[\frac{1}{2^2}-\frac{1}{\infin^2}]$
$\frac{1}{λ}=\frac{R}{4}\ \ ....(i)$
The shortest wavelength in the Brackett series occurs when an electron transitions from $\infty\ \text{to}\ n = 4$.
$\frac{1}{\lambda}=R(1)^2[\frac{1}{4^2}-\frac{1}{∞^2}]$
$⇒\frac{1}{λ'}=\frac{R}{16}\ \ ...(ii)$
Now, By dividing Equation (i) and Equation (ii)
$\frac{λ'}{λ}=\frac{R}{4}\times\frac{16}{R}⇒λ'=4λ$
So, the correct option is (C) : 4λ.