If an electron in a hydrogen atom jumps from the 3^rd orbit to the 2^nd orbit, it emits a photon of wavelength λ. When it jumps from the 4^th orbit to the 3^rd orbit, the corresponding wavelength of the photon will be

Question

Question

Answer 1

When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,

\(\frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right)\)

So, \(\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) = \frac{{5R}}{{36}}\)

and \(\frac{1}{{\lambda '}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) = \frac{{7R}}{{144}}\)

∴ \(\lambda ' = \frac{{144}}{7} \times \frac{{5\lambda }}{{36}} = \frac{{20\lambda }}{7}\)