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Chapter 36: Q. 38 (page 1060)

At what speed, as a fraction of $c$ , must an electron move so that its total energy is $10 %$ more than its rest mass energy?

Short Answer

Expert verified

$v = 1.24 脳 10^{8} m / s$

Step by step solution

01

Given Information

We have given that, $c = 3 脳 10^{8} m / s$

Total energy is $10 %$ more.

02

Simplify

Given values are: $c = 3 脳 10^{8} m / s$ and total energy is $10 %$ more.

To find the speed $v$ , we use the below given formula.

role="math" localid="1649826808978" $\frac{m c^{2}}{\sqrt{(1 - \frac{v^{2}}{c^{2}})}} = m c^{2} + 10 % of m c^{2}$

Now, put the given values in the above formula to find the speed.

$\frac{m c^{2}}{\sqrt{(1 - \frac{v^{2}}{c^{2}})}} = m c^{2} + 0.1 脳 m c^{2} \frac{m c^{2}}{\sqrt{(1 - \frac{v^{2}}{c^{2}})}} = 1.1 m c^{2} \frac{1}{\sqrt{(1 - \frac{v^{2}}{c^{2}})}} = 1.1 \sqrt{(1 - \frac{v^{2}}{c^{2}})} = \frac{1}{1.1} 1 - \frac{v^{2}}{c^{2}} = \frac{1}{1.21} \frac{v^{2}}{c^{2}} = \frac{0.21}{1.21} v^{2} = c^{2} 路 \frac{21}{121} v = c 路 \sqrt{\frac{21}{121}} v = 1.24 脳 10^{8} m / s$

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