Question

Calculate the charge on an alpha particle .given charge on a proton is equal to 1.6 x10 ^ -19 C

Answer 1

Hope this helps comment if you want further explanation

Answer 2

The charge of an alpha particle is $\bold{2.032 \times 10^{-13} \ \mathrm{m}}$

Given:

Charge on proton =1.6 \times 10^{-19} \ C

Solution:

We know that,

$1 \ \mathrm{KeV}=1000 \ \mathrm{eV}$

$1 \ e V=1.6 \times 10^{-19} \ \mathrm{J}$

The particle's Kinetic energy which is given here is,

$1.6 \times 10^{-19} \times 10^{4} \ J=1.6 \times 10^{-15} \ J$

Kinetic energy is,

$K E=\frac{1}{2} m v^{2}$

$\text{Alpha particle's mass} =4 \times 1.67 \times 10^{-27} \ k g=6.68 \times 10^{-27} \ k g$

$\Rightarrow K E=\frac{1}{2} m v^{2}$

$\Rightarrow 1.6 \times 10^{-15}=6.68 \times 10^{-27} \times v^{2}$

$\Rightarrow v^{2}=\frac{1.6 \times 10^{-15}}{6.68 \times 10^{-27}}$

$\Rightarrow v^{2}=2.39 \times 10^{11}$

$v=4.88 \times 10^{5} \ m s^{-1}$

The Plank's constant, $h=6.626 \times 10^{-34} \ m^2.kg/s$

According to de-Broglie equation,

$\Rightarrow \lambda=\frac{h}{m v}$

Here, h is defined as plank's constant.

$\Rightarrow \frac{6.626 \times 10^{-34}}{6.68 \times 10^{-27} \times 4.88 \times 10^{5}}$

$\Rightarrow \lambda=2.032 \times 10^{-13} \mathrm{m}$