Question

A uniform magnetic field with a slit system is shown in the figure. This setup is to be used as a momentum fliter for high energy charged particles. With a field of $B\text{T}$, it is found that the filter transmits $\alpha$-particle each of energy <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $5.3\text{MeV}$. The magnetic field is increased to $2.3B\text{T}$ and deutrons are passed into the filter. what is the energy of each deutron transmitted by filter?

• A. $4.5\text{MeV}$
• B. $8.6\text{MeV}$
• C. $14.02\text{MeV}$
• D. $18.04\text{MeV}$

Answer 1

The correct option is C $14.02\text{MeV}$

In case of a circular motion of charged particle in magnetic field the K.E of the particle is,

$K.E=\frac{r^2{q}^2{B}^2}{2m}$

$(\because r=\frac{mv}{qB}=\frac{m\sqrt{\frac{2KE}{m}}}{qB})$

According to question $\alpha$ -particle gets filtered with energy <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $5.3\text{MeV}$.

$(K.E{)}_{\alpha }=\frac{r^2\left(2e{)}^2\right(B{)}^2}{2\left(4m\right)}$

Where, $m$ is mass of the proton.

$(K.E{)}_D=\frac{r^2\left(e{)}^2\right(2.3B{)}^2}{2\left(2m\right)}$

$\Rightarrow \frac{(K.E{)}_D}{(K.E{)}_{\alpha }}=\frac{(2.3{)}^2\times 4}{2^2\times 2}$

or,

$(K.E{)}_D=(K.E{)}_{\alpha }\times \left(\frac{5.29}{2}\right)=\left(5.3\right)\times \left(\frac{5.29}{2}\right)$

$\therefore (K.E{)}_D\simeq 14.02\text{MeV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(c\right)$ is correct.