Dual Nature Of Light Question 67
Question: In a Thomson set-up for the determination of e/m, electrons accelerated by 2.5 $ kV $ enter the region of crossed electric and magnetic fields of strengths $ 3.6\times 10^{4}V{{m}^{-1}} $ and $ 1.2\times {{10}^{-3}}T $ respectively and go through undeflected. The measured value of $ e/m $ of the electron is equal to [AMU 2002]
Options:
A) $ 1.0\times 10^{11}C\text{-}k{{g}^{-1}} $
B) $ 1.76\times 10^{11}C\text{-}k{{g}^{-1}} $
C) $ 1.80\times 10^{11}C\text{-}k{{g}^{-1}} $
D) $ 1.85\times 10^{11}C\text{-}k{{g}^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{e}{m}=\frac{E^{2}}{2VB^{2}}=\frac{{{(3.6\times 10^{4})}^{2}}}{2\times 2.5\times 10^{3}\times {{(1.2\times {{10}^{-3}})}^{2}}} $ $ =1.8\times 10^{11}C/kg $ .
BETA
