Nuclear Physics And Radioactivity Question 110
Question: In an experiment for positive ray analysis with Thomson method, two identical parabola are obtained when applied electric fields are 3000 and 2000 V/m. The particles are singly ionised particles assuming same magnetic field:
Options:
A) 1 : 3
B) 2 : 4
C) 3 : 1
D) 4 : 2
Show Answer
Answer:
Correct Answer: A
Solution:
[a] For same magnetic field $ \frac{y^{2}}{x}\propto \frac{1}{E}( \frac{e}{m} ) $ For singly ionized particle, $ \frac{y^{2}}{x}\propto \frac{1}{E_{1}}\frac{e}{m_{1}} $ For doubly ionised particle, $ \frac{y^{2}}{x}\propto \frac{1}{E_{2}}\frac{2e}{m_{2}} $ Since the parabolas for both the particles are identical, $ \frac{1}{E_{1}}\frac{e}{m_{1}}=\frac{1}{E_{2}}\frac{2e}{m_{2}} $ So $ \frac{m_{1}}{m_{2}}=\frac{E_{2}}{2E_{1}}=\frac{2000}{2\times 3000}=\frac{1}{3} $
BETA
