Magnetic Effects Of Current Question 351
Question: A proton of energy 8 eV is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be [J & K CET 2004]
Options:
A) 4 eV
B) 2 eV
C) 8 eV
D) 6 eV
Show Answer
Answer:
Correct Answer: C
Solution:
$ r=\frac{\sqrt{2mK}}{qB}\Rightarrow q\propto \sqrt{mK}\Rightarrow K\propto \frac{q^{2}}{m} $
$ \Rightarrow \frac{{K_{\alpha }}}{K_{p}}={{( \frac{{q_{\alpha }}}{q_{p}} )}^{2}}\times \frac{m_{p}}{{m_{\alpha }}}\Rightarrow \frac{{K_{\alpha }}}{8}={{( \frac{2q_{p}}{q_{p}} )}^{2}}\times \frac{m_{p}}{4m_{p}}=1 $
$ \Rightarrow {K_{\alpha }}=8\ eV $
BETA
