Question

The shortest wavelength in the Balmer series is $(R = 1.097 \times 10^7 \,m^{-1})$

• A. $200\, nm $
• B. $256.8\, nm $
• C. $300\, nm $
• D. $364.6\, nm $

Answer 1

Answer: (D)

Wavelength for Balmer series is
$ \frac{1}{\lambda} = R \left(\frac{1}{2^{2}} -\frac{1}{n^{2}}\right) $
Here, Rydberg's constant
$R = 1.097 \times 10^7\, m^{-1}$
$\therefore \frac{1}{\lambda} = \frac{4}{1.097\times10^{7} } $
$= 364.6 \times 10^{-9} m $
$= 364.6 \,nm$