First slide

Force on a charged particle moving in a magnetic field

Question

A proton of energy 8 e V 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

Moderate

A

4 eV
B

2 eV
C

8 eV
D

6 eV

Solution

$\begin{array}{l} r = \frac{\sqrt{2 mK}}{qB} \Rightarrow q \propto \sqrt{mK} \Rightarrow K \propto \frac{q^{2}}{m} \\ \Rightarrow \frac{K_{α}}{K_{p}} = {(\frac{q_{α}}{q_{p}})}^{2} \times \frac{m_{p}}{m_{α}} \Rightarrow \frac{K_{α}}{8} = {(\frac{2 q_{p}}{q_{p}})}^{2} \times \frac{m_{p}}{4 m_{p}} = 1 \\ \Rightarrow K_{α} = 8 eV \end{array}$

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