Electromagnetic Waves - Result Question 24
24. If $\lambda_v, \lambda_x$ and $\lambda_m$ represent the wavelengths of visible light, $X$-rays and microwaves respectively, then
(a) $\lambda_m>\lambda_x>\lambda_v$
(b) $\lambda_m>\lambda_v>\lambda_x$
(c) $\lambda_v>\lambda_x>\lambda_m$
(d) $\lambda_v>\lambda_m>\lambda_x$
[2005]
Show Answer
Answer:
Correct Answer: 24. (b)
Solution:
- (b) We know $E=\frac{h c}{\lambda} \Rightarrow E \propto \frac{1}{\lambda}$
$\Rightarrow E_m<E_v<E_x$
$\therefore \lambda_m>\lambda_v>\lambda_x$
Energy of electromagnetic wave, $E=\frac{h c}{\lambda}$
For $e m$ wave of lower energy, $(v)$ is small and $\lambda$ is larger. For $e m$ wave of higher energy (v) is large and $\lambda$ is small.
