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2. Electric Potential and Capacitance
easy
The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100\, volts$ is
A
$1$
B
$\sqrt {\frac{{2{m_e}}}{{{m_\alpha }}}} $
C
$\sqrt {\frac{{{m_e}}}{{{m_\alpha }}}} $
D
$\sqrt {\frac{{{m_e}}}{{2{m_\alpha }}}} $
Solution
(d) Momentum $p = \sqrt {2mK} \,;$ where $K =$ kinetic energy = $Q.V$
$==>$ $p = \sqrt {2mQV} $ $==>$ $p \propto \sqrt {mQ} \, \Rightarrow \,\frac{{{p_e}}}{{{p_\alpha }}} = \sqrt {\frac{{{m_e}{Q_e}}}{{{m_\alpha }{Q_\alpha }}}} $$ = \sqrt {\frac{{{m_e}}}{{2{m_\alpha }}}} $
Standard 12
Physics
