Question

Calculate the amount of work done to increase the speed of an electron from 0.6c to 0.8c . Given that the rest mass energy of electron is 0.511 MeV​

Answer 1

Answer:

3.82×10^(-23)joule

Explanation:

Mass of electron = 9.1 × 10^-23 kilograms

Work done = change in energy(kinetic + potential)

But nothing said about potential so assume potential energy is not changing

Kinetic energy =1/2 × m × (v2^2 - v1^2)

V2 = final velocity

V1 = initial velocity

So work done = 1/2 × 9.1 × 10^(-31) × (0.8c^2-0.6c^2) joule

On solve the above expression we will get the result

3.82×10^(-23)joule

Answer 2

Answer:

The amount of work done to increase the speed of an electron from 0.6c to 0.8c is equal to 0.215MeV.

Explanation:

We know that the work done is equal to change in the kinetic energy.

For relative particle (electron):-

Total energy = Kinetic energy + rest mass energy

Kinetic energy = mc² - m₀c²

$m=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2} } }$

Where m₀ is the rest mass of electron, m is relativistic mass of electron  and c is speed of light.

Given, the rest mass energy of electron, m₀c² = 0.511 MeV​

The initial speed of electron = 0.6c

The final speed of electron = 0.8c

The change in kinetic energy, ΔK.E. is given by:

$\triangle K.E.=\big [\frac{m_oc^2}{\sqrt{1-\frac{(0.8)^2c^2}{c^2} } }-m_oc^2\big ] - \big [\frac{m_oc^2}{\sqrt{1-\frac{(0.6)^2c^2}{c^2} } }-m_oc^2\big ]$

$\triangle K.E.=m_oc^2\big [\frac{1}{\sqrt{1-0.64 } }- \big [\frac{1}{\sqrt{1-0.36} } ]$

$\triangle K.E.=m_oc^2\big [\frac{1}{\sqrt{0.36} }- \frac{1}{\sqrt{0.64} } ]$

$\triangle K.E.=m_oc^2\big [\frac{1}{0.6 }- \frac{1}{0.8} ]$

$\triangle K.E.=m_oc^2\big [1.67 -1.25 ]$

$\triangle K.E.=0.511\times 0.42\; MeV$

$\triangle K.E.=0.215\;MeV$

Therefore, the amount of work done is equal to 0.215MeV.

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