The ratio of specific charge of an alpha -pa | vedclass

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Question

(c) Specific charge $ = \frac{q}{m}$; Ratio $ = \frac{{{{\left( {\frac{q}{m}} \right)}_\alpha }}}{{{{\left( {\frac{q}{m}} \right)}_p}}} = \frac{{{q_\a

Vedclass · Accepted Answer

$1 : 2$

Vedclass · Answer

(c) Specific charge $ = \frac{q}{m}$; Ratio $ = \frac{{{{\left( {\frac{q}{m}} ight)}_\alpha }}}{{{{\left( {\frac{q}{m}} ight)}_p}}} = \frac{{{q_\alpha }}}{{{q_p}}} 	imes \frac{{{m_p}}}{{{m_\alpha }}} = \frac{1}{2}$.

The ratio of specific charge of an $\alpha $ -particle to that of a proton is

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