Question

Calculate the shortest wavelength of electromagnetic radiation present in Balmer series of hydrogen spectrum.

Answer 1

Wave length of photo emitted in balmer series of hydrogen atom
$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2})$
where $n=3,4,\mathrm{5......}$
for minimum wavelength $n=\infty$
So,$\frac{1}{{\lambda }_{min}}=R(\frac{1}{2^2}-\frac{1}{\infty })$
$=\frac{R}{4}$
${\lambda }_{min}=\frac{4}{R}=\frac{4}{1.097\times {10}^{-7}}$
$364.6nm$