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

Question

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

Tardigrade Admin · Accepted Answer

(1/2) m v2 =e V ⇒ v =√(2 e V/m) p =m v=√2 m e V Now, pe =√2 me × e × 100 and pα=√4 mα × e × 100 ∴ (pe/pα) =√(me/2 mα)

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

$1$

$(2 m_{e} / m_{a})$

$(m_{e} / m_{α})$

$(m_{e} /2 m_{α})$

$\frac{1}{2} m v^{2} = e V$
$\Rightarrow v = \frac{2 e V}{m}$
$p = m v = 2 m e V$
Now, $p_{e} = 2 m_{e} \times e \times 100$
and $p_{α} = 4 m_{α} \times e \times 100$
$∴ \frac{p _{e}}{p _{α}} = \frac{m _{e}}{2 m _{α}}$

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

1

(2me​/ma​)​

(me​/mα​)​

(me​/2mα​)​

21​mv2=eV ⇒v=m2eV​​ p=mv=2meV​ Now, pe​=2me​×e×100​ and pα​=4mα​×e×100​ ∴pα​pe​​=2mα​me​​​

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

$1$

$(2 m_{e} / m_{a})$

$(m_{e} / m_{α})$

$(m_{e} /2 m_{α})$

$\frac{1}{2} m v^{2} = e V$
$\Rightarrow v = \frac{2 e V}{m}$
$p = m v = 2 m e V$
Now, $p_{e} = 2 m_{e} \times e \times 100$
and $p_{α} = 4 m_{α} \times e \times 100$
$∴ \frac{p _{e}}{p _{α}} = \frac{m _{e}}{2 m _{α}}$