JEE Main 12 Jan 2019 Morning Question 30
Question: A particle A of mass m and charge q is accelerated by a potential difference of 50 V. Another particle B of mass 4 m and charge q is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths $ \frac{{\lambda_A}}{{\lambda_B}} $ is close to [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 14.14
B) 0.07
C) 4.47
D) 10.00
Show Answer
Answer:
Correct Answer: A
Solution:
For particle A $ m _1=m,q _1=q;V _1=50,V $ $ {\lambda_A}=\frac{h}{\sqrt{2m _1q _1V _1}} $ …(i)
For particle B $ m _2=4m,q _2=q;V _2=2500,V $ $ {\lambda_B}=\frac{h}{\sqrt{2m _2q _2V _2}} $ …(ii)
Dividing eqn. (i) by (ii),
$ \frac{{\lambda_A}}{{\lambda_B}}=\sqrt{\frac{m _2q _2v _2}{m _1q _1v _1}}=\sqrt{\frac{4m\times q\times 2500}{m\times q\times 50}} $ $ \frac{{\lambda_A}}{{\lambda_B}}=\sqrt{\frac{4\times 2500}{50}}=14.14 $
