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

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(C) Given that the kinetic energy $K_p = K_{\alpha}$.We know that the radius of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}

Vedclass · Accepted Answer

The $\alpha$-particle and the proton will be bent in a circular path with the same radius.

Vedclass · Answer

(C) Given that the kinetic energy $K_p = K_{\alpha}$.We know that the radius of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.For the proton: $r_p = \frac{\sqrt{2m_p K_p}}{q_p B}$.For the $\alpha$-particle: $r_{\alpha} = \frac{\sqrt{2m_{\alpha} K_{\alpha}}}{q_{\alpha} B}$.Taking the ratio: $\frac{r_p}{r_{\alpha}} = \sqrt{\frac{m_p}{m_{\alpha}}} 	imes \frac{q_{\alpha}}{q_p} = \sqrt{\frac{m}{4m}} 	imes \frac{2e}{e} = \frac{1}{2} 	imes 2 = 1$.Therefore,$r_p = r_{\alpha}$. Both particles will move in a circular path with the same radius.

$A$ 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 a uniform magnetic field. Which one of the following statements is true?

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