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Arlecino [84]
3 years ago
8

Determine the energy required to accelerate an electron from (a) 0.500c to 0.900c and (b) 0.900c to 0.990c

Physics
1 answer:
Lera25 [3.4K]3 years ago
6 0

Answer:

0.582 MeV

2.45 MeV

Explanation:

E_r=0.511\ MeV=Electron rest energy

(a) 0.500c to 0.900c

W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.9^2}}-\frac{1}{\sqrt{1-0.5^2}}\right)0.511\\\Rightarrow W=0.582\ MeV

Energy required is 0.582 MeV

(b) 0.900c to 0.990c

W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.99^2}}-\frac{1}{\sqrt{1-0.9^2}}\right)0.511\\\Rightarrow W=2.45\ MeV

Energy required is 2.45 MeV

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(c)

E' = P'/t'............. Equation 3

E = P/t................ Equation 4

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From the question,

Given: t' = 15 minutes = (15×60) = 900 seconds, t = 30 minutes = (30×60) = 1800 seconds

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As the coiled spring experiences a compression force and is then released, it experiences a sequential movement of the wave of the compression that extends the length of the coiled spring which is then followed by a stretched section of the coiled spring in a repeatedly such that the direction of vibration of particles of the coiled is parallel to direction of motion of the wave

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