91Ó°ÊÓ

Initial velocity, ${\mathrm{v}}_0=1.50×{10}^5\mathrm{m}/\mathrm{s}$

Final velocity, $\mathrm{v}=5.70×{10}^5\mathrm{m}/\mathrm{s}$

Distance travelled, x=1.00 cm

Acceleration can be found by using the given data in the kinematic equation which involves initial velocity, final velocity, acceleration, and displacement terms.

The formula to find acceleration can be obtained with the help of final as well as initial velocity.

$$\begin{gathered} v^2=v_0^2+2ax \\ \mathrm{Here},\mathrm{v}\mathrm{is}\mathrm{the}\mathrm{final}\mathrm{velocity},{\mathrm{v}}_0\mathrm{is}\mathrm{the}\mathrm{initial}\mathrm{velocity},\mathrm{a}\mathrm{is}\mathrm{the}\mathrm{acceleration}\mathrm{and}\mathrm{x}\mathrm{is}\mathrm{the}\mathrm{distance}\mathrm{travelled}. \end{gathered}$$

Convert 1.00 cm into i.e. in S.I. units.

$$\begin{gathered} 1\mathrm{cm}=\frac{1\mathrm{cm}}{100\mathrm{cm}}×1\mathrm{m} \\ =0.01m \\ \mathrm{Rearrange}\mathrm{the}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{acceleration}\mathrm{as}, \\ \mathrm{a}=\frac{{\mathrm{v}}^2-{\mathrm{v}}_0^2}{2\mathrm{x}} \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ \mathrm{a}=\frac{{\left(5.7×{10}^6\mathrm{m}/\mathrm{s}\right)}^2-{\left(1.5×{10}^5\mathrm{m}/\mathrm{s}\right)}^2}{2×0.01\mathrm{m}} \\ =1.62×{10}^{15}\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Therefore},\mathrm{the}\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{electron}\mathrm{to}\mathrm{be}\mathrm{assumed}\mathrm{constant}\mathrm{is}1.62×{10}^{15}\mathrm{m}/{\mathrm{s}}^2. \end{gathered}$$