Question

In hydrogen spectrum, the shortest wavelength in Balmer series is l. The shortest wavelength in Brackett series is –

• A. 2l
• B. 4l
• C. 9l
• D. 16l

Answer 1

Answer: (B)

4l

Answer 2

$\frac{1}{\lambda} = R\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$

for shortest wavelength of Balmer series

n₁ = 2 ; n₂ = 

$\frac{1}{\lambda} = R\left\lbrack \frac{1}{4} - \frac{1}{\infty} \right\rbrack \Rightarrow \frac{4}{R}$

$\frac{1}{\lambda'} = R\left\lbrack \frac{1}{4^{2}} - \frac{1}{\infty} \right\rbrack$

\ l' = $\frac{16}{R} = 4 \times \frac{4}{R}$ = 4l