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Q87E

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Modern Physics
Found in: Page 68
Modern Physics

Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

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Short Answer

A proton is accelerated from rest through a potential difference of 500 MV.

(a) What is its speed?

(b) Classical mechanics indicates that quadrupling the potential difference would double the speed. Were a classical analysis valid, what speed would result from a 2000 MV potential difference?

(c) What speed actually results?

  1. At 500MV, speed of a proton is 0.7567c.
  2. At 2000MV, classical speed is 2c
  3. The relativistic speed is 0.9476c.
See the step by step solution

Step by Step Solution

Step 1: Determine relation between kinetic energy and potential difference

To find the speed of a proton accelerated under a potential difference of 500MV, we must equate relativistic kinetic energy to change in potential energy.

Relativistic kinetic energy is equal to change in potential energy as the charged particle is accelerated under a potential difference

localid="1659086760380" KE=UKE=eV ...................(1)

Here, KE is the kinetic energy, ΔU is the change in potential energy, e is the charge, and ΔV is the potential difference.

And as the electron is traveling at a very high speed considering relativistic Kinetic energy as below.

KE= Total energy - moc2 = mc2 - moc2= γmoc2 - moc2

KE= γ - 1moc2 ….. (2)

Here, m0 is the rest mass which is equal to the mass of the proton. Therefore, by comparing equation (1) and (2), you get

γ - 1mpc2 = eΔV ….. (3)

Step 2: (a) Speed when potential difference is500MV.

Consider the known data as below.

The mass of proton, mp=1.67 × 10-27 kg

The speed of light, c=3 × 108 ms

The charge, e=1.6 × 10-19 C

The potential difference, ΔV=500 MV=5× 108 V

Substitute all known numerical values into equation (3), and you have

(γ-1)(1.67×10-27kg)(3×108ms)2=(1.6×10-19C)(5×108V) (γ-1)=0.532 γ=1.532

Then using formula of Lorentz factor

localid="1659087345441" γ=11-v2c21.532=11-v2c21-v2c2=12.347v2c2=1 - 0.426v2=0.574c2v=0.758c

Hence, at 500MV , the peed of a proton is 0.758c.

Step 3: (b) Classical speed when potential difference is quadrupled that is V=2000MV.

Here, we will equate classical kinetic energy to potential energy

12mv2=eV v2=2eVm v2=21.6×10-19C2×1091.67×10-27 v2c

Hence, at ,2000MV the classical speed is 2c .

According to the Einstein’s second postulate that, the speed of light is constant irrespective of the reference frame.

Therefore, classical result is invalid.

Step 4: (c) Determine the relativistic speed to get the actual result:Rewrite equation (3) as below.

γ - 1mpc2 = eΔV

Here,

The mass of proton, mp=1.67 × 10-27 kg

The speed of light, c=3 × 108 ms

The charge, e=1.6 × 10-19 C

The potential difference, ΔV=2000 MV=2× 109 V

Substitute all known numerical values in the above equation.

γ - 1 1.67 × 10-27 kg 3 × 108 ms2=1.6 × 10-19 C2× 109 Vγ - 1 15.03 × 10-11= 3.2× 10-10γ - 1 = 2.129γ = 3.129

Then using formula of Lorentz factor as given by,

γ=11-v2c23.129 = 11-v2c21-v2c2=19.791v2c2= 1 - 0.102v2=0.898c2v=0.947c

Hence, at 2000MV , the Speed of a proton is 0.947c.

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