Question

A proton and alpha particle are initially at distance'r' apart .Find the kinetic energy of proton at large seperation from alpha particle after being released

Answer 1

Given :

Initial distance between the proton and the alpha particle = r

To Find :

What is the  kinetic energy of proton at large separation from the alpha particle after being released ?

Solution :

Let mass of proton  (   $H^+$  ) = m

Then the mass of alpha particle (  $He^{2+$)  will be = 4m

Potential energy of the system is :

=$\frac{K \times e \times 2e}{r}$        Here , e is charge on $H^+$ and 2e is charge on $He^{2+$

Now by using the law of conservation of energy :

Final K.E of system = Initial P.E

Or , $\frac{1}{2}mv_1^2 + \frac{1}{2}4mv_2^2 = \frac{K.e.2e}{r}$  

∴No external force acts on the system , the centre of mass of the system has zero velocity ( centre of mass will not shift ) .

Now taking A as origin :

$(C.O.M )_x = \frac{m(0) + 4m(r)}{m + 4m } = \frac{4r}{5}$

Also ,  $(C.O.M )_x = \frac{m(x_1) + 4m(x_2)}{m + 4m }$

Or , $(C.O.M )_x = \frac{x_1}{5} + \frac{4x_2}{5}$

So , $V_{C.O.M} = \frac{V_1}{5} + \frac{4V_2}{5}$

Or , $0 = \frac{V_1}{5} + \frac{4V_2}{5}$

Or , $V_1 = -4V_2$

Or , $V_2^2 = \frac{V_1^2}{16}$

So we have :

$\frac{1}{2}mV_1^2 + \frac{1}{2}4m \times \frac[V_1^2}{16} = \frac{k.2e^2}{r}$

Or ,$\frac{1}{2}mV_1^2[1 + \frac{4}{16}] = \frac{k.2e^2}{r}$

Or , $\frac{1}{2}mV_1^2 = \frac{k.2e^2}{r}\times \frac{16}{20}$

                 = $\frac{8}{5} \frac{ke^2}{r}$

Hence , the  Kinetic energy of proton at large separation is   $\frac{8}{5} \frac{ke^2}{r}$ .