A proton (mass m and charge +e) and an alpha& | vedclass

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Question

Given than $K_{p}=K_{a}$

We know that $r=\frac{m v}{q B}=\frac{(2 m K)^{1 / 2}}{q B}$

$	herefore r_{p}=\frac{\left(2 m_{p} K_{p}ight)^{1 / 2}

Vedclass · Accepted Answer

The radius of the path of the $\alpha  -$ particle will be greater than that of the proton

Vedclass · Answer

Given than $K_{p}=K_{a}$

We know that $r=\frac{m v}{q B}=\frac{(2 m K)^{1 / 2}}{q B}$

$	herefore r_{p}=\frac{\left(2 m_{p} K_{p}ight)^{1 / 2}}{q_{p} B}$ and $r_{\alpha}=\frac{\left(2 m_{\alpha} K_{\alpha}ight)^{1 / 2}}{q_{\alpha} B}$

Now $\frac{r_{p}}{r_{\alpha}}=\sqrt{\left(\frac{m_{p}}{m_{\alpha}}ight)} 	imes \frac{q_{\alpha}}{q_{p}}=\sqrt{\left(\frac{m_{p}}{4 m_{p}}ight)} 	imes \frac{2 e}{e}=1$

A proton (mass $m$ and charge $+e$) and an $\alpha -$ particle (mass $4m$ and charge $+2e$) are projected with the same kinetic energy at right angles to the uniform magnetic field. Which one of the following statements will be true

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