Question

if an alpha particle and a proton are accelerated from rest by a potential difference of 1 megavolt then ratio of their kinetic energy

Answer 1

2:1 I Think because kinetic energy 1/2mv^2= Vq here V=voltage q =charge and charge of alpha particle is twice the proton so it should be 2/1

Answer 2

The ratio of kinetic energy is 1 : 1

Given:

Potential difference = 1 megavolt

Solution:

The kinetic energy will have a ratio calculated through the formula of kinetic energy that is,

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

And the Force is given by the formulas

F = q E and F = m a

On equating both formulas, we get,

$qE = ma$

$a=\frac{q E}{m}$

Thereby, velocity after time t will be, $v = u + a t$

Since, particle starts from rest, $v = a t$

$v=\frac{q E}{m} \times t$

Now, K.E of proton is given below:

$K \cdot E_{p}=\frac{1}{2} m\left(\frac{q E t}{m}\right)^{2}$

$\Rightarrow K . E_{p}=\frac{1}{2}\left(\frac{q^{2} E^{2} t^{2}}{m}\right)$

Alpha particle has 4m mass and charge of 2q.  

$F = 2 q E$

Now, K.E. of alpha is given below:

$K . E_{a}=\frac{1}{2} \times 4 m \times\left(\frac{2 q E}{4 m} \times t\right)^{2}$

$K . E_{a}=\frac{1}{2} \times 4 m\left(\frac{q^{2} E^{2} t^{2}}{4 m^{2}}\right)$

$\Rightarrow K \cdot E_{a}=\frac{1}{2}\left(\frac{q^{2} E^{2} t^{2}}{m}\right)$

Since, the Kinetic energy of alpha and proton is same, so their ratio be 1 : 1