The ratio of momenta of an electron and an α -particle which are accelerated from rest by a potential difference of 100 V is

Question

The ratio of momenta of an electron and an α -particle which are accelerated from rest by a potential difference of 100 V is (A)  1  (B)  √(2me/mα )  (C)  √(me/mα )  (D)  √(me/2mα )

Tardigrade Admin · Accepted Answer

Momentum, p=mv and F=√(2QV/m) ⇒ p=√2QmV p propto √Qm ∴ (pe/pα )=√(eme/2emα )=√(me/2mα )

Q. The ratio of momenta of an electron and an $α$ -particle which are accelerated from rest by a potential difference of 100 V is

$1$

$\frac{2 m _{e}}{m _{α}}$

$\frac{m _{e}}{m _{α}}$

$\frac{m _{e}}{2 m _{α}}$

Momentum, $p = m v$
and $F = \frac{2 Q V}{m}$
$\Rightarrow$ $p = 2 Q mV$
$p \propto Q m$
$∴$ $\frac{p _{e}}{p _{α}} = \frac{e m _{e}}{2 e m _{α}} = \frac{m _{e}}{2 m _{α}}$

Q. The ratio of momenta of an electron and an α -particle which are accelerated from rest by a potential difference of 100 V is

1

mα​2me​​​

mα​me​​​

2mα​me​​​

Momentum, p=mv and F=m2QV​​ ⇒ p=2QmV​ p∝Qm​ ∴ pα​pe​​=2emα​eme​​​=2mα​me​​​

Q. The ratio of momenta of an electron and an $α$ -particle which are accelerated from rest by a potential difference of 100 V is

$1$

$\frac{2 m _{e}}{m _{α}}$

$\frac{m _{e}}{m _{α}}$

$\frac{m _{e}}{2 m _{α}}$

Momentum, $p = m v$
and $F = \frac{2 Q V}{m}$
$\Rightarrow$ $p = 2 Q mV$
$p \propto Q m$
$∴$ $\frac{p _{e}}{p _{α}} = \frac{e m _{e}}{2 e m _{α}} = \frac{m _{e}}{2 m _{α}}$