Question

Calculate the ratio of kinetic energy of alpha particles and proton when they are placed in electric field E.

Answer 1

Step 1: Given that:

Alpha particle and proton are placed in a uniform electric field.

Step 2: Formula used:

Charge on proton = $+e$

Alpha particle = Nucleus of helium

Charge on alpha particle = $+2e$

If the mass of the proton is $m$ then mass of alpha particle is $4m$ .

The force on a charge q in a uniform electric field E is given by

$F=qE$

But, F= ma ( By second law of motion)

Thus,

$ma=qE$

$a=\frac{qE}{m}$

The initial velocity of proton, as well as alpha particle, is taken as zero.

According to the first equation of motion, $v=u+at$

Thus, $v=at$

The kinetic energy of a body of mass m moving with velocity v is given by $\frac{1}{2}m{v}^2$ .

Step 3: Calculation of velocity of alpha particle and proton:

Acceleration in proton($a_p$) = $\frac{eE}{m}$

Acceleration in alpha particle($a_{he}$) = $\frac{2eE}{m}$

Let the velocity of proton is $v_p$ and the velocity of alpha particle is $v_{he}$ then

Using the first equation of motion

For time t;

$v_p=\frac{eE}{m}\times t$

And for alpha particle;

$v_{he}=\frac{2eE}{m}\times t$

Step 4: Calculation of ratio of the kinetic energy of an alpha particle to proton:

The kinetic energy of the proton

$K_p=\frac{1}{2}\times m\times {\left(\frac{eE}{m}t\right)}^2$

$K_p=\frac{1}{2m}{\left(eEt\right)}^2$ .............(1)

The kinetic energy of the alpha particle will be;

$K_{\alpha }=\frac{1}{2}\times 4m\times {(\frac{2eE}{m}\times t)}^2$

$K_{\alpha }=2m\times 4\frac{{\left(eEt\right)}^2}{m^2}$

$K_{\alpha }=8\frac{{\left(eEt\right)}^2}{m}$.............(2)

Now, dividing equation (2) by equation 1), we get;

$\frac{K_{\alpha }}{K_p}=\frac{8\frac{{\left(eEt\right)}^2}{m}}{\frac{1}{2}\frac{{\left(eEt\right)}^2}{m}}$

$\frac{K_{\alpha }}{K_p}=\frac{16}{1}$

Thus,

The ratio of the kinetic energy of an alpha particle to the kinetic energy of the proton is $16:1$.