Question

An electron in an atom jumping from $3^{rd}$ orbit to $2^{nd}$ orbit emits radiation of wavelength ${\lambda }_1$ and when it jumps from $2^{nd}$ orbit to $1^{st}$ orbit emits radiation of wavelength ${\lambda }_2$. The wavelength of radiation emitted when it jumps from $3^{rd}$ orbit to $1^{st}$ is

• A. $\sqrt{{\lambda }_1{\lambda }_2}$
• B. $\frac{{\lambda }_1+{\lambda }_2}{2}$
• C. $\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$
• D. ${\lambda }_1+{\lambda }_2$

Answer 1

The correct option is C $\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$

When an electron jumps from $n_y$th orbit to $n_x$th orbit , the wavelength $\lambda$ of emitted radiation is given by ,

$\frac{1}{\lambda }=Z^{2}R[1/n_x^2-1/n_y^2]$

Hence , when an electron jumps from $3rd$ orbit to $2nd$ orbit , the wavelength ${\lambda }_1$ of emitted radiation will be ,

$\frac{1}{{\lambda }_1}=Z^{2}R[1/2^2-1/3^2]$ ............eq1

and , when an electron jumps from $2nd$ orbit to $1st$ orbit , the wavelength ${\lambda }_2$ of emitted radiation will be ,

$\frac{1}{{\lambda }_2}=Z^{2}R[1/1^2-1/2^2]$ ...............eq2

Adding eq1 and eq2 , we get

$\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=Z^{2}R[1/2^2-1/3^2+1/1^2-1/2^2]$

or $\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=Z^{2}R[1/1^2-1/3^2]$

or $\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1{\lambda }_2}=Z^{2}R[1/1^2-1/3^2]$ .............eq3

Now , when an electron jumps from $3rd$ orbit to $1st$ orbit , the wavelength ${\lambda }_3$ of emitted radiation will be ,

$\frac{1}{{\lambda }_3}=Z^{2}R[1/1^2-1/3^2]$ ...............eq4

comparing eq3 and eq4 , we get

$\frac{1}{{\lambda }_3}=\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1{\lambda }_2}$

or ${\lambda }_3=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$