A Van de Graaff accelerator utilizes a\(50.0{\rm{ }}MV\)potential difference to accelerate charged particles such as protons.

(a) What is the velocity of a proton accelerated by such a potential?

(b) An electron?

The value of potential difference of Van de Graff accelerator is:

\(50.0{\rm{ }}MV\)

In essence, relativistic kinetic energy describes a particle's kinetic energy as the difference between that energy and its rest mass energy. This equation resembles the non-relativistic kinetic energy statement for low velocities.

With the help of the equation (\({\rm{28}}{\rm{.52}}\)), the relativistic kinetic energy is obtained by:

\(K{\rm{ }}{E_{rel}}{\rm{ }} = {\rm{ }}(\gamma {\rm{ }} - {\rm{ }}1){\rm{ }}m{c^2}\)…………….(I)

We have the value of \(\gamma \) as the relativistic constant.

The value of \(m{c^2}\) is said to be the rest mass energy of the particle.

The charged particles of the kinetic energy then is:

\(K{E_{rel}} = (\gamma - 1){\rm{ }}m{\rm{ }}{c^2}\)…………….(II)

The value of \(e\) is the electron charge.

The value of \(V\) is the applied voltage to accelerate charged particles.

The relativistic factor is then obtained as:

\(\gamma = \dfrac{1}{{\sqrt {1 - {{(\dfrac{u}{c})}^2}} }}\)…………….(III)

The value of \(u\) is the speed of the object.

With the help of the first and second equations, we get:

\(eV = (\gamma - 1)m{c^2}\)

OR,\(\gamma - 1 = \dfrac{{eV}}{{m{c^2}}}\)

OR, \(\gamma = \dfrac{{eV}}{{m{c^2}}} + 1\)…………….(IV)

Now with the help of the third and fourth equations, we get:

\(\dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{u}{c}} \right)}^2}} }} = \dfrac{{eV}}{{m{c^2}}} + 1\)

OR, \(\dfrac{1}{{1 - {{\left( {\dfrac{u}{c}} \right)}^2}}} = {\left( {\dfrac{{eV}}{{m{c^2}}} + 1} \right)^2}\)

OR, \(1 - {\left( {\dfrac{u}{c}} \right)^2} = \dfrac{1}{{{{\left( {\dfrac{{eV}}{{m{c^2}}} + 1} \right)}^2}}}\)

OR, \({\left( {\dfrac{u}{c}} \right)^2} = 1 - \dfrac{1}{{{{\left( {\dfrac{{eV}}{{m{c^2}}} + 1} \right)}^2}}}\)

OR, \(\begin{aligned}u &= c\sqrt {1 - \dfrac{1}{{{{\left( {\dfrac{{eV}}{{m{c^2}}} + 1} \right)}^2}}}} \\ &= c\sqrt {1 - \dfrac{1}{{{{\left( {\dfrac{{eV}}{{m{c^2}}} + 1} \right)}^2}}}} .\end{aligned}\)…………….(V)

Here it is given that the accelerating potential is:

\(\begin{aligned}V{\rm{ }} &= {\rm{ }}50.0{\rm{ }}MV\\ &= {\rm{ }}50.0 \times {10^6}{\rm{ }}V\end{aligned}\)

The mass of the proton is:

\({m_p} = 1.67 \times {10^{ - 27}}{\rm{ }}kg\)

The mass of the electron is:

\({m_e} = 9.11 \times {10^{ - 31}}{\rm{ }}kg\)

The rest mass energy of the proton then will be:

\({(m{\rm{ }}{c^2})_p}{\rm{ }} = {\rm{ }}938.3{\rm{ }}MeV\)

The rest mass energy of the electron then will be:

\({(m{\rm{ }}{c^2})_e}{\rm{ }} = {\rm{ }}0.511{\rm{ }}MeV\)

\(\begin{aligned}{u_p}{\rm{ }} &= c\sqrt {1 - \dfrac{1}{{{{\left( {\dfrac{{eV}}{{{m_p}{c^2}}} + 1} \right)}^2}}}} \\ &= c\sqrt {1 - \dfrac{1}{{{{\left( {\dfrac{{1.6 \times {{10}^{ - 19}}C \times 50.0 \times {{10}^6}\;V}}{{1.67 \times {{10}^{27}}\;kg \times {{\left( {3.00 \times {{10}^8}\;m\;{s^{ - 1}}} \right)}^2}}} + 1} \right)}^2}}}} \\ &= 0.379576c\\ &= 0.380c\end{aligned}\)

Therefore, the velocity of proton is: \(0.380{\rm{ }}c\).

\(\begin{aligned}{u_e}{\rm{ }} &= c\sqrt {1 - \dfrac{1}{{{{\left( {\dfrac{{eV}}{{{m_e}{c^2}}} + 1} \right)}^2}}}} \\ &= c\sqrt {1 - \dfrac{1}{{{{\left( {\dfrac{{1.6 \times {{10}^{19}}C \times 50.0 \times {{10}^6}\;V}}{{9.11 \times {{10}^{31}}\;kg \times {{\left( {3.00 \times {{10}^8}\;m{s^{ - 1}}} \right)}^2}}} + 1} \right)}^2}}}} \\& = 0.99995c\end{aligned}\)

Therefore, the velocity of electron is: \(0.99995{\rm{ }}c\).