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

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(c) $r = \frac{{\sqrt {2mK} }}{{qB}}i.e.\;r \propto \frac{{\sqrt m }}{q}$
Here kinetic energy $K$ and $B$ are same.
$	herefore \frac{{{r_p}}}{{{r_\

Vedclass · Accepted Answer

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

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(c) $r = \frac{{\sqrt {2mK} }}{{qB}}i.e.\;r \propto \frac{{\sqrt m }}{q}$
Here kinetic energy $K$ and $B$ are same.
$	herefore \frac{{{r_p}}}{{{r_\alpha }}} = \frac{{\sqrt {{m_p}} }}{{\sqrt {{m_\alpha }} }}.\frac{{{q_\alpha }}}{{{q_p}}} = \frac{{\sqrt {{m_p}} }}{{\sqrt {4{m_p}} }}.\frac{{2{q_p}}}{{{q_p}}} = 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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