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A proton and a deuteron are both accelerated through the same potential difference and enter in a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius $R$, assuming the neutron and proton masses are nearly equal, the radius of the proton's path will be
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The correct answer is:
$\frac{R}{\sqrt{2}}$
$\frac{R}{\sqrt{2}}$
As charge on both proton and deuteron is same i.e. ' $e$ '
Energy acquired by both, $E=\mathrm{eV}$ For Deuteron.
Kinetic energy, $\frac{1}{2} m V^2=e V$
[ $V$ is the potential difference]
$$
v=\sqrt{\frac{2 e V}{m_d}}
$$
But $m_d=2 m$
Therefore, $v=\sqrt{\frac{2 e V}{2 m}}=\sqrt{\frac{e V}{m}}$
Radius of path, $R=\frac{m v}{e B}$
Substituting value of ' $v$ ' we get
$$
\begin{array}{r}
R=\frac{2 m \sqrt{\frac{e v}{m}}}{e B} \\
\frac{R}{2}=\frac{m \sqrt{\frac{e v}{m}}}{e B}
\end{array}
$$
For proton :
$$
\begin{aligned}
& \frac{1}{2} m V^2=e V \\
& V=\sqrt{\frac{2 e V}{m}} \\
& \text { Radius of path, } R^{\prime}=\frac{m V}{e B}=\frac{m \sqrt{\frac{2 e V}{m}}}{e B} \\
& R^{\prime}=\sqrt{2} \times \frac{R}{2} \quad \text { [From eq. (i)] } \\
& R^{\prime}=\frac{R}{\sqrt{2}}
\end{aligned}
$$
Energy acquired by both, $E=\mathrm{eV}$ For Deuteron.
Kinetic energy, $\frac{1}{2} m V^2=e V$
[ $V$ is the potential difference]
$$
v=\sqrt{\frac{2 e V}{m_d}}
$$
But $m_d=2 m$
Therefore, $v=\sqrt{\frac{2 e V}{2 m}}=\sqrt{\frac{e V}{m}}$
Radius of path, $R=\frac{m v}{e B}$
Substituting value of ' $v$ ' we get
$$
\begin{array}{r}
R=\frac{2 m \sqrt{\frac{e v}{m}}}{e B} \\
\frac{R}{2}=\frac{m \sqrt{\frac{e v}{m}}}{e B}
\end{array}
$$
For proton :
$$
\begin{aligned}
& \frac{1}{2} m V^2=e V \\
& V=\sqrt{\frac{2 e V}{m}} \\
& \text { Radius of path, } R^{\prime}=\frac{m V}{e B}=\frac{m \sqrt{\frac{2 e V}{m}}}{e B} \\
& R^{\prime}=\sqrt{2} \times \frac{R}{2} \quad \text { [From eq. (i)] } \\
& R^{\prime}=\frac{R}{\sqrt{2}}
\end{aligned}
$$
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