A proton and a deuteron having same kinetic energies enter a region of uniform magnetic field , moving perpendicular to . The ratio of the radius of deuteron path to the radius of the proton path is:

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Solution

Q.
A proton and a deuteron $(q = + e, m = 2.0 u)$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$ , moving perpendicular to $\vec{B}$ . The ratio of the radius $r_{d}$ of deuteron path to the radius $r_{p}$ of the proton path is: [2024]

1 $1 : 1$

2 $1 : \sqrt{2}$

3 $\sqrt{2} : 1$

4 $1 : 2$

Ans.
(3)

$As radius of circular path (r) = \frac{m v}{q B}$ $\dots (i)$

$Given {(K E)}_{p} = {(K E)}_{d}$

$\frac{1}{2} m_{p} v_{p}^{2} = \frac{1}{2} m_{d} v_{d}^{2} \Rightarrow m_{p} v_{p}^{2} = m_{d} v_{d}^{2}$

$As m_{d} = 2 m_{p} \Rightarrow m_{p} v_{p}^{2} = 2 m_{p} v_{d}^{2} \Rightarrow v_{p} = \sqrt{2} v_{d}$

Now from equation (i),

$\Rightarrow \frac{r_{p}}{r_{d}} = \frac{m_{p} v_{p}}{e B} \times \frac{e B}{2 m_{p} v_{d}} = \frac{1}{\sqrt{2}} = r_{d} : r_{p} = \sqrt{2} : 1$

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