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Q23P

Expert-verifiedFound in: Page 33

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An electron with an initial velocity${{\mathbf{\text{V}}}}_{{\mathbf{0}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{50}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$****enters a region of length L=1.00cm**** where it is electrically accelerated. It emerges with${\mathbf{v}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{70}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{6}}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$****. What is its acceleration, assumed constant?**

The acceleration of electron to be assumed constant is $1.62\times {10}^{15}\mathrm{m}/{\mathrm{s}}^{2}$ .

Initial velocity, ${\mathrm{v}}_{0}=1.50\times {10}^{5}\mathrm{m}/\mathrm{s}$

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

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.

${v}^{2}={v}_{0}^{2}+2ax\phantom{\rule{0ex}{0ex}}\mathrm{Here},\mathrm{v}\mathrm{is}\mathrm{the}\mathrm{final}\mathrm{velocity},{\mathrm{v}}_{0}\mathrm{is}\mathrm{the}\mathrm{initial}\mathrm{velocity},\mathrm{a}\mathrm{is}\mathrm{the}\mathrm{acceleration}\mathrm{and}\mathrm{x}\mathrm{is}\mathrm{the}\mathrm{distance}\mathrm{travelled}.$

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

$1\mathrm{cm}=\frac{1\mathrm{cm}}{100\mathrm{cm}}\times 1\mathrm{m}\phantom{\rule{0ex}{0ex}}=0.01m\phantom{\rule{0ex}{0ex}}\mathrm{Rearrange}\mathrm{the}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{acceleration}\mathrm{as},\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{{\mathrm{v}}^{2}-{\mathrm{v}}_{0}^{2}}{2\mathrm{x}}\phantom{\rule{0ex}{0ex}}\mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}.\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{{\left(5.7\times {10}^{6}\mathrm{m}/\mathrm{s}\right)}^{2}-{\left(1.5\times {10}^{5}\mathrm{m}/\mathrm{s}\right)}^{2}}{2\times 0.01\mathrm{m}}\phantom{\rule{0ex}{0ex}}=1.62\times {10}^{15}\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{electron}\mathrm{to}\mathrm{be}\mathrm{assumed}\mathrm{constant}\mathrm{is}1.62\times {10}^{15}\mathrm{m}/{\mathrm{s}}^{2}.$

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