91影视

Consider the formula for the work required to move a charge from the point a to the point b is:

${\mathbf{W}}_{\mathbf{a}\mathbf{鈫�}\mathbf{b}}\mathbf{=}{\mathbf{U}}_{\mathbf{b}}\mathbf{-}{\mathbf{U}}_{\mathbf{a}}$

Here,$U_a$and$U_b$are the initial and final electric potential respectively.

Consider the expression for the initial potential as:

$U_a=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{q_1{q}_2}{r_a}$

Consider the expression for the final potential as:

$U_b=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{q_2{q}_2}{r_b}$

Consider the formula for the change in the kinetic energy when the mass of proton and electron are equal and are projected in the same direction.

$$\begin{gathered} 鈭�K=\frac{m{v}_2^2}{2}+\frac{m{v}_2^2}{2} \\ =m{v}_2^2 \end{gathered}$$

Then, the change in the work done is.

$鈭�U=m{v}_2^2$

Rewrite the expression in terms of the velocity as,

$v_2=\sqrt{\frac{鈭�U}{m}}$ 鈥�.. (1)

(a)

Consider the equation for the work done as:

$$\begin{gathered} W=U_b-U_a \\ W=\frac{1}{4{\mathrm{蟺蔚}}_0}\left[\frac{1}{r_b}-\frac{1}{r_a}\right] \end{gathered}$$

Put given values

$$\begin{gathered} \mathrm{W}=\left(9脳{10}^9\right){\left(1.6脳{10}^{-19}\right)}^2\left[\frac{1}{3.0脳{10}^{-15}}-\frac{1}{2脳{10}^{-10}}\right] \\ \mathrm{W}=-7.68脳{10}^{-14}\mathrm{J} \end{gathered}$$

But work done is never negative so$7.68脳{10}^{-14}鈥�\mathrm{J}$.

Hence, the amount done is $7.68脳{10}^{-14}鈥�\mathrm{J}$.

(b)

Substitute the values in the equation (1) and solve as:

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{7.68脳{10}^{-14}\mathrm{J}}{1.67脳{10}^{-27}\mathrm{kg}}} \\ \mathrm{v}=6.78脳{10}^6\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, the velocity is $6.78脳{10}^6\frac{\mathrm{m}}{\mathrm{s}}$.