Question

A proton, a direction ion and an $\alpha$-particle of equal kinetic energy perform circular motion normal to a uniform magnetic field $B$. If the radii of their paths are $r_p,r_d$ and $r_{\alpha }$ respectively then [Here, $q_d=q_p,m_d=2m_p$]

• A. $r_{\alpha }=r_p>r_d$
• B. $r_{\alpha }=r_p<r_d$
• C. $r_{\alpha }>r_d>r_p$
• D. $r_{\alpha }=r_d=r_p$

Answer 1

Answer: (B)

$r_{\alpha }=r_p<r_d$

$\begin{array}{c}\text{Given,}\\ \text{a proton}q=+1m=m_p\\ \text{a deutorium}q=+1m=2m_p\\ \text{a}\alpha \text{particle}q=+2m=4m_p\end{array}$

$\begin{array}{c}\text{hane equal Kinctic energy}\\ B=\text{magnetic field.}\end{array}$

$\begin{array}{c}\text{We Know when a change particle enters}\\ \text{in a magnetic field it perform}\\ \text{circular motion with radius equal to.}\end{array}$

$\begin{array}{cc}r& =\frac{mv}{qB}\\ r& =\frac{\sqrt{2mKE}}{qB}\end{array}$

$\begin{array}{c}{r}_{\text{proton}}=\frac{\sqrt{2\times m_p\times KE}}{1\times B}\\ {\text{r}}_{deutorium}=\frac{\sqrt{2\times 2\times m_p\times KE}}{1\times B}\\ {\text{r}}_{\alpha }=\frac{\sqrt{2\times 4\times m_p\times KE}}{2\times B}=\frac{\sqrt{2\times m_p\times KE}}{1\times B}\end{array}$

$\text{So,}{r}_{\text{proton}}=r_{\alpha }<r_{\text{deutorium}}\left(\text{option}B\right)$