Question

Energy levels A, B and C of a certain atom correspond to increasing values of energy, i.e., $E_A<E_B<E_C$. If ${\lambda }_1$, ${\lambda }_2$ and ${\lambda }_3$ are the wavelengths of the radiation corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct?

• A. ${\lambda }_3={\lambda }_1+{\lambda }_2$
• B. ${\lambda }_3=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$
• C. ${\lambda }_1+{\lambda }_2+{\lambda }_3=0$
• D. ${\lambda }_3^2={\lambda }_1^2+{\lambda }_2^2$

Answer 1

The correct option is B ${\lambda }_3=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$

When an electron falls from energy level $E_2$ to energy level $E_1$, the frequency $v$ of the emitted radiation is given by
$hv=E_2-E_1$
$\Rightarrow \frac{hc}{\lambda }=E_2-E_1$
$\Rightarrow \frac{1}{\lambda }=\frac{1}{hc}(E_2-E_1)=k(E_2-E_1)$ where $k=\frac{1}{hc}$
For energy levels A, B and C, we have
$\frac{1}{{\lambda }_1}=k(E_C-E_B)$.....(i)
$\frac{1}{{\lambda }_2}=k(E_B-E_A)$.....(ii)
and $\frac{1}{{\lambda }_3}=k(E_C-E_A)$.....(iii)
Adding (i), (ii) and (iii),
$\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}+\frac{1}{{\lambda }_3}=k\left\{\right(E_C-E_B)+(E_B-E_A)+(E_C-E_A\left)\right\}=2k(E_C-E_A)=\frac{2}{{\lambda }_3}\Rightarrow \frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}\Rightarrow {\lambda }_3=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$
Hence, the correct choice is (b).