Question

A deuteron of kinetic energy 50keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of proton that describes a circular orbit of radius 0.5 metre in the same plane with same B is

• A. 25 KeV
• B. 50 KeV
• C. 100 KeV
• D. 200 KeV

Answer 1

The correct option is C 100 KeV

For deuteron;Given, kinetic energy$=50Kev$

radios of circular orbit =0.5m=r$

For proton;kinetic energy$=?$

radios of circular orbit $=0.5m=r$

we have to find kinetic energy of proton.

for a charge particle orbiting in a circular path in a magnetic field.

$\frac{m{v}^2}{r}=qvb$

$\Rightarrow v=\frac{qbr}{m}$

B=magnetic field.

q=charge on a particle

m=mass of a particle (proton)

r=radius of circular orbit

v=velocity of charge particle.

kinetic energy;$E_k=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{qbr}{m}\right)}^2=\frac{q^2{r}^2{B}^2}{2m}fordeutron,E_1=\frac{B^2{q}^2\times r^2}{2\times 2m}$

as mass of deuteron$=2\times$mass of proton.

for proton,$E_2=\frac{B^2{q}^2\times r^2}{2m}$

dividing $E_{1}by{E}_2$,we get

$\frac{E_1}{E_2}=\frac{1}{2}\Rightarrow E_2=2\times E_1=2\times 50kev=100kev$

Hence the answer is option (c).