2022-01-15

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

encolatgehu

Expert

Step 1
$\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$
$K=\frac{m{c}^{2}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}-m{c}^{2}=\left(\gamma -1\right)m{c}^{2}$
$E=K+m{c}^{2}$, (E is the total energy of a praticle)
Step 2
a)
$\frac{m{c}^{2}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}-m{c}^{2}=7.50×{10}^{5}×1.602×{10}^{-19}J=1.2016×{10}^{-13}J$
$m=9.1×{10}^{-31}kg$
$\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}=\frac{m{c}^{2}}{1.2016×{10}^{-13}J+m{c}^{2}}=0.4052$
$\frac{{v}^{2}}{{c}^{2}}=1-0.4052$
$v=c\sqrt{1-0.4052}=0.9142c$
Step 3
b) $k=\frac{1}{2}m{v}^{2}$
$v=\sqrt{\frac{2k}{m}}=\sqrt{\frac{2×1.2016×{10}^{-13}J}{9.1×{10}^{-31}kg}}=5.13×{10}^{8}m/s$

otoplilp1

Expert

Step 1
Given;
potential of the electron, $V=750kV$
kinetic energy of the electron, $K.E=7.50×{10}^{5}eV$
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
$k=E-m{c}^{2}$
$k=\frac{m{c}^{2}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}-m{c}^{2}=7,5×{10}^{5}×1.6×{10}^{-19}J=12×{10}^{-14}J$
$\frac{m{c}^{2}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}=12×{10}^{-14}J+m{c}^{2}$
$\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}=\frac{m{c}^{2}}{12×{10}^{-14}J+m{c}^{2}}$
but, mass of electron, $m=9.1×{10}^{-31}kg$ and speed of light $c=3×{10}^{8}\frac{m}{s}$
$\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}=\frac{9.1×{1.0}^{-31}{\left(3×{10}^{8}\right)}^{2}}{12×{10}^{-14}+9.1×{10}^{-31}{\left(3×{10}^{8}\right)}^{2}}=0.4057$
$1-\frac{{v}^{2}}{{c}^{2}}={0.4057}^{2}$
$1-\frac{{v}^{2}}{{c}^{2}}=0.1646$
$\frac{{v}^{2}}{{c}^{2}}=1-0.1646$
$\frac{{v}^{2}}{{c}^{2}}=0.8354$
$\frac{v}{c}=0.914$
$v=0.914c$
Step 2
b) speed of the electron when computed from principles of classical mechanics
$k=\frac{1}{2}m{v}^{2}$
${v}^{2}=\frac{2k}{m}$
$v=\sqrt{\frac{2k}{m}}$
$v=\sqrt{\frac{2×12×{10}^{-14}}{9.1×{10}^{-31}}}$

alenahelenash

Expert

Step 1
The formula to calculate relativistic kinetic energy of the particle is,
$K=\gamma m{c}^{2}-m{c}^{2}$
m is mass of the electron
$m{c}^{2}$ is rest mass energy of the electron
$\gamma$ is the relativistic factor,
c is the speed of light,
The relativistic factor $\gamma$ is,
$\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$
Use $\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$ for $\gamma$ in the force equation to rewrite K,
$K=\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}m{c}^{2}-m{c}^{2}$
rewrite the above expression in terms of v.
$\frac{m{c}^{2}}{\sqrt{1-\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}}}=K+m{c}^{2}$
$1-\frac{{v}^{2}}{{c}^{2}}=\frac{m{c}^{2}}{K+m{c}^{2}}$
$v=c\sqrt{1-\frac{m{c}^{2}}{K+m{c}^{2}}}$
Substitute $9.1×{10}^{-31}kg$ for m, $3×{10}^{8}m{s}^{-1}$ for c and $7.50×{10}^{5}eV$ for K in the above velocity equation to find v,
$v=c\sqrt{1-\frac{\left(9.1×{10}^{-31}kg\right)\left(3×{10}^{8}m×{s}^{-1}{\right)}^{2}}{\left(7.50×{10}^{5}eV×\left(\frac{1.6×{10}^{-19}J}{cV}\right)\right)+\left(9.1×{10}^{-31}kg\right)\left(3×{10}^{8}m×{s}^{-1}{\right)}^{2}}}$
$=c\sqrt{1-0.128}$
$=0.934c$
Therefore, the velocity of the electron is 0.934c
Step 2
b) Kinetic energy of the electron K is $7.50×{10}^{5}eV$
The formula to calculate the classical kinetic energy of the electron is given by,
$K=\frac{1}{2}m{v}^{2}$
From the above non relativistic kinetic enrgy equation velocity of the electron is,
$\frac{2K}{m}={v}^{2}$
$v=\sqrt{\frac{2K}{m}}$
Substitute $7.50×{10}^{5}eV$ for K in the above velocity equation to find v.
$v=\sqrt{\frac{2\left(7.50×{10}^{5}eV×\left(\frac{1.6×{10}^{-19}J}{cV}\right)\right)}{\left(9.1×{10}^{-31}kg\right)}}$
$=\sqrt{2.64×{10}^{17}{m}^{2}×{s}^{-2}}$
$=5.14×{10}^{8}m×{s}^{-1}$

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