Question
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $$\lambda .$$ When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be
A.
$$\frac{{16}}{{25}}\lambda $$
B.
$$\frac{9}{{16}}\lambda $$
C.
$$\frac{{20}}{7}\lambda $$
D.
$$\frac{{20}}{{13}}\lambda $$
Answer :
$$\frac{{20}}{7}\lambda $$
Solution :
Excess energy of $${e^ - }$$ appears as photon.
From Rydberg’s formula,
$$\eqalign{
& \frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right) = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) = \frac{{5R}}{{36}} \cr
& \frac{1}{{\lambda '}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) = \frac{{7R}}{{144}} \cr
& \frac{{\frac{1}{\lambda }}}{{\frac{1}{{\lambda '}}}} = \frac{{5R}}{{36}} \div \frac{{7R}}{{144}} \cr
& \Rightarrow \frac{{\lambda '}}{\lambda } = \frac{{5R}}{{36}} \times \frac{{144}}{{7R}} = \frac{{20}}{7} \cr
& \Rightarrow \lambda ' = \frac{{20}}{7}\lambda \cr} $$