Question

The shortest wavelength in the Balmer series is

$(\text{Take }R = 1.097 \times 10^{7}\text{ }\text{m}^{- 1})$

• A. 200 nm
• B. 256.8 nm
• C. 300 nm
• D. 364.6 nm

Answer 1

Answer: (D)

364.6 nm

Answer 2

Wavelength of Balmer series is

$\frac{1}{\lambda} = R\left( \frac{1}{2^{2}} - \frac{1}{n^{2}} \right)$

At $n = \infty$, the limit of the series observed

$\therefore\frac{1}{\lambda} = R\left( \frac{1}{4} - \frac{1}{\infty^{2}} \right)$

$\frac{1}{\lambda} = \frac{R}{4}or\lambda = \frac{4}{R}$

Here, Rydberg’s constant

$R = 1.097 \times 10^{7}m^{- 1}$

$\therefore\lambda = \frac{4}{1.097 \times 10^{7}}$

$= 364.6 \times 10^{- 9}m = 364.6nm$