(a) What voltage will accelerate electrons to a speed of\({\rm{6}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{m/s}}\)?

(b) Find the radius of curvature of the path of a proton accelerated through this potential in a \({\rm{0}}{\rm{.500 - T}}\)field and compare this with the radius of curvature of an electron accelerated through the same potential.

Kinetic energy is the energy that an item or particle has as a result of its movement. When work is done on an object by exerting a net force, the object accelerates and gains kinetic energy as a result.

Speed of the electrons gained \( = v = 6 \times {10^{ - 7}}{\rm}\) m/s

Voltage through which the electrons accelerate\( = \Delta V\)

Charge on the electron\(q = 1.6 \times {10^{ - 19}}{\rm}\)

Mass of the electron\({{\rm{m}}_{\rm{e}}}{\rm{ = 9}}{\rm{.1 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}{\rm{ kg}}\)

Kinetic energy is given as

\(KE = \frac{1}{2}m{v^2}\)

Electric potential energy is given as

\(U = q\Delta V\)

Speed of the proton=\({V_P}\)

Voltage through which the proton is accelerated\( = \Delta v\)=\(1.02 \times {10^{ - 24}}\;{\rm{V}}\)

Charge on the proton\(q = 1.6 \times {10^{ - 19}}{\rm}\)

Mass of the proton

The radius of curvature for proton\( = {r_p}\)

The radius of curvature of a charged particle is given as

\(r = \frac{{mv}}{{qB}}\)

(a)

The kinetic energy obtained by the electron equals the electric potential energy lost by the electron, therefore

\(\begin{aligned}KE = U\\\frac{1}{2}m{v^2} = q\Delta V\end{aligned}\)

Substitute the values

\(\Delta V = {\rm{1}}{\rm{.02}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 24}}}}\;{\rm{V}}\)

Thus, the voltage through which the electron is accelerated comes out to be \({\rm{1}}{\rm{.02}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 24}}}}\;{\rm{V}}\)

(b)

The kinetic energy obtained by the electron equals the electric potential energy lost by the electron, therefore

\(\begin{aligned}KE = U(0.5)\\{m_p}v_p^2 = q\Delta V\end{aligned}\)

Substitute the values

\(\begin{aligned}\frac{1}{2}\left( {1.67 \times {{10}^{ - 27}}\;{\rm{kg}}} \right){v_p^2} &= \left( {1.6 \times {{10}^{ - 19}}\;{\rm}} \right)\left( {1.02 \times {{10}^{ - 24}}\;{\rm{V}}} \right)\\{v_p} &= 1.4 \times {10^{ - 8}}{\rm}\ \end{aligned}\)

The radius of curvature of the proton is given as

\({r_p} = \frac{{{m_p}{v_p}}}{{qB}}\)

Substitute the values

\({{\rm{r}}_{\rm{p}}}{\rm{ = }}\frac{{{\rm{2 \times }}\left( {{\rm{1}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}} \right)}}{{\left( {{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right)}}\)

\({{\rm{r}}_{\rm{p}}}{\rm{ = 2}}{\rm{.92 \times 1}}{{\rm{0}}^{{\rm{ - 16}}}}{\rm{ m}}\)

The radius of curvature of the electron is given as

\({r_e} = \frac{{{m_e}{v_e}}}{{qB}}\)

Substitute the values

\({{\rm{r}}_{\rm{e}}}{\rm{ = }}\frac{{\left( {{\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}\;{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{6 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}\;{\rm{m}}} \right)}}{{\left( {{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right){\rm{ \times (0}}{\rm{.5)}}}}\)

\({{\rm{r}}_{\rm{e}}}{\rm{ = 6}}{\rm{.83 \times 1}}{{\rm{0}}^{{\rm{ - 18}}}}{\rm{ m}}\)

The ratio of the radius of the proton to the electron is given as

\(\begin{aligned}{\rm{ratio }} &= \frac{{{r_p}}}{{{r_e}}}\\{\rm{ratio}} &= \frac{{\left( {2.92 \times {{10}^{ - 16}}\;{\rm{m}}} \right)}}{{\left( {6.83 \times {{10}^{ - 18}}\;{\rm{m}}} \right)}}\\{\rm{ratio}} &= 43 \end{aligned}\)

Therefore, the radius of curvature of the proton comes out to be and it is 43 times that of the radius of curvature of the electron.