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

Question

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 (A) √2R (B) (R/√2) (C) (R/2) (D) R

Tardigrade Admin · Accepted Answer

As charge on both proton and deuteron is same i.e.

^{'} e^{'}

Energy acquired by both,

E = e V

For Deuteron.
Kinetic energy,

\frac{1}{2} m V^{2} - e V

[

V

is the potential difference]

v = \frac{2 e V}{m _{d}}

But

m_{d} = 2 m

Therefore,

v = \frac{2 e V}{2 m} = \frac{e V}{m}

Radius of path,

R = \frac{m v}{e B}

Substituting value of

^{'} v^{'}

we get

R = \frac{2 m \frac{e v}{m}}{e B}

\frac{R}{2} = \frac{m \frac{e v}{m}}{e B} ... (i)

For proton :

\frac{1}{2} m V^{2} = e V

V = \frac{2 e V}{m}

Radius of path,

R^{'} = \frac{mV}{e B} = \frac{m \frac{2 e V}{m}}{e B}

R^{'} = 2 \times \frac{R}{2}

[From eq. (i)]

R^{'} = \frac{R}{2}

$2 R$

$\frac{R}{2}$

$\frac{R}{2}$

$R$

2​R

2​R​

2R​

R

$2 R$

$\frac{R}{2}$

$\frac{R}{2}$

$R$