Work Power Energy Question 4
Question 4 - 2024 (29 Jan Shift 1)
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25 \%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:
(1) $\frac{1}{1}$
(2) $\frac{1}{8}$
(3) $\frac{8}{1}$
(4) $\frac{1}{4}$
Show Answer
Answer: (2)
Solution:
For photon
$E _p=\frac{hc}{\lambda _p} \Rightarrow \lambda _p=\frac{hc}{E _p}$
For electron
$\lambda _e=\frac{h}{m _e v _e}=\frac{hv _e}{2 K _e}$
Given $v _c=0.25 c$
$\lambda _e=\frac{h \times 0.25 c}{2 K _e}=\frac{hc}{8 K _e}$
Also $\lambda _p=\lambda _e$
$\frac{hc}{E _p}=\frac{hc}{8 K _e}$
$\frac{K _e}{E _p}=\frac{1}{8}$
