maduregimc

Answered

2022-01-15

Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy is $\frac{7}{50}\times {10}^{5}eV$

a) What is the ratio of the speed v of an electron having this energy to the speed of light, c?

b) What would the speed be if it were computed from the principles of classical mechanics?

a) What is the ratio of the speed v of an electron having this energy to the speed of light, c?

b) What would the speed be if it were computed from the principles of classical mechanics?

Answer & Explanation

encolatgehu

Expert

2022-01-16Added 27 answers

Step 1

Step 2

a)

Step 3

b)

otoplilp1

Expert

2022-01-17Added 41 answers

Step 1

Given;

potential of the electron,$V=750kV$

kinetic energy of the electron,$K.E=7.50\times {10}^{5}eV$

Part (A) the ratio of the speed of an electron having this energy to the speed of light

$E=k+m{c}^{2}$

where;

E is the total energy of the electron

k is kinetic energy

$k=E-m{c}^{2}$

$k=\frac{m{c}^{2}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}-m{c}^{2}=7,5\times {10}^{5}\times 1.6\times {10}^{-19}J=12\times {10}^{-14}J$

$\frac{m{c}^{2}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}=12\times {10}^{-14}J+m{c}^{2}$

$\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}=\frac{m{c}^{2}}{12\times {10}^{-14}J+m{c}^{2}}$

but, mass of electron,$m=9.1\times {10}^{-31}kg$ and speed of light $c=3\times {10}^{8}\frac{m}{s}$

$\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}=\frac{9.1\times {1.0}^{-31}{(3\times {10}^{8})}^{2}}{12\times {10}^{-14}+9.1\times {10}^{-31}{(3\times {10}^{8})}^{2}}=0.4057$

$1-\frac{{v}^{2}}{{c}^{2}}={0.4057}^{2}$

$1-\frac{{v}^{2}}{{c}^{2}}=0.1646$

$\frac{{v}^{2}}{{c}^{2}}=1-0.1646$

$\frac{{v}^{2}}{{c}^{2}}=0.8354$

$\frac{v}{c}=0.914$

$v=0.914c$

Step 2

b) speed of the electron when computed from principles of classical mechanics

$k=\frac{1}{2}m{v}^{2}$

$v}^{2}=\frac{2k}{m$

$v=\sqrt{\frac{2k}{m}}$

$v=\sqrt{\frac{2\times 12\times {10}^{-14}}{9.1\times {10}^{-31}}}$

Given;

potential of the electron,

kinetic energy of the electron,

Part (A) the ratio of the speed of an electron having this energy to the speed of light

where;

E is the total energy of the electron

k is kinetic energy

but, mass of electron,

Step 2

b) speed of the electron when computed from principles of classical mechanics

alenahelenash

Expert

2022-01-23Added 366 answers

Step 1

The formula to calculate relativistic kinetic energy of the particle is,

m is mass of the electron

c is the speed of light,

The relativistic factor

Use

rewrite the above expression in terms of v.

Substitute

Therefore, the velocity of the electron is 0.934c

Step 2

b) Kinetic energy of the electron K is

The formula to calculate the classical kinetic energy of the electron is given by,

From the above non relativistic kinetic enrgy equation velocity of the electron is,

Substitute

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