Question

In the balmer series of atomic spectra of hydrogen atom, a line corresponding to wavelength $656.4nm$ was obtained. Calculate the no. of higher orbit from which the electron drops to produce this line.

Answer 1

In Balmer series, transmission of electron is from $nth$ state to $n=2$ state.

Wavelength $\lambda =656.4nm$

So for Balmer series, $\frac{1}{\lambda }=(1.097\times {10}^7)(\frac{1}{4}-\frac{1}{n^2})$

Or $\frac{1}{656.4\times {10}^{-9}}=(1.097\times {10}^7)(\frac{1}{4}-\frac{1}{n^2})$

$⟹n=3$