Dual Nature of Radiation and Matter - Result Question 72
76. Energy levels $A, B, C$ of a certain atom correspond to increasing values of energy i.e., $E_A<E_B<E_C$ If $\lambda_1, \lambda_2, \lambda_3$ are the wavelengths of radiation corresponding to the transitions $C$ to $B, B$ to $A$ and $C$ to $A$ respectively, which of the following relation is correct?
[1990, 2005]
(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) $\lambda_3{ }^{2}=\lambda_1{ }^{2}+\lambda_2{ }^{2}$
Show Answer
Answer:
Correct Answer: 76. (b)
Solution:
- (b) $(E_2-E_1)=h v=\frac{h c}{\lambda}$
$\therefore \frac{h c}{\lambda_1}=(E_C-E_B), \frac{h c}{\lambda_2}=(E_B-E_A)$
and $\frac{h c}{\lambda_3}=(E_C-E_A)$
Now,
$(E_C-E_A)=(E_C-E_B)+(E_B-E_A)$
or, $\frac{h c}{\lambda_3}=\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2}$ or $\frac{1}{\lambda_3}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$
$\therefore \frac{1}{\lambda_3}=\frac{\lambda_1+\lambda_2}{\lambda_1 \lambda_2}$ or $\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}$
