Question

A steady beam of alpha particles (q2e) traveling with constant kinetic energy 20 MeV carries a current of0.25 A . (a) If the beam is directed perpendicular to a flat surface, how many alpha particles strike the surface in 3.0 s? (b) At any instant, how many alpha particles are there in a given 20 cmlength of the beam? (c) Through what potential difference is it necessary to accelerate each alpha particle from rest to bring it to energy of 20 MeV ?

Short answer

1. The number of alpha particles that strikes the surface in if the beam is directed perpendicular to the flat surface is $2.3\times {10}^{12}$.
2. The number of alpha particles present in a given 20 cm length of the beam is $5\times {10}^3$.
3. The potential difference that is necessary to accelerate each alpha particle from rest to bring it to energy of 20 MeV is $1\times {10}^7\mathrm{V}$.

Step-by-step solution

The given data:

Constant kinetic energy of the beam, K = 20 Mev

Current of the particles,$\mathrm{i}=0.25\mathrm{\mu A}$

The steady beam of alpha particles is directed perpendicular to the flat surface.

Time of the strike of the particles, t = 3 s

Length of the beam, L = 20 cm

Understanding the concept of the current and potential difference:

An alpha particle consists of protons and neutrons, so the charge on an alpha particle is . To find the alpha particle number, first find the moving total charge. Dividing this total charge by the charge of the alpha particle gives the number of moving alpha particles. Using the kinetic energy of a particle, define the velocity of the particle, and then calculate the number of alpha particles in a given beam. Use the law of conservation of energy to calculate the potential difference to accelerate the alpha particle.

Formulas:

The charge contained for the flow of current is given by,

$∆\mathrm{q}=\mathrm{i}∆\mathrm{t}$ ….. (1)

The number of particles contained within the charge is as below.

$\mathrm{N}=∆\mathrm{q}/\mathrm{e}$ ….. (2)

The kinetic energy of the particle in motion is,

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$ ….. (3)

The law of conservation of the energy gives is define by using following formula.

${\mathrm{K}}_{\mathrm{i}}+{\mathrm{U}}_{\mathrm{i}}={\mathrm{K}}_{\mathrm{f}}+{\mathrm{U}}_{\mathrm{f}}$ ….. (4)

(a) Calculation of the number of alpha particles that strike the surface:

The total charge that strikes the surface can be calculated by substituting known values into equation (1) as follows:

$$\begin{gathered} ∆\mathrm{q}=0.25\times {10}^{-6}\mathrm{A}\times 3\mathrm{s} \\ =0.75\times {10}^{-6}\mathrm{C} \end{gathered}$$

Since, each particle carries charge 2e the number of particle that strikes the surface is given using equation (2) as follows: (charge on one proton, $\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$).

role="math" localid="1661833502645" $$\begin{gathered} \mathrm{N}=\frac{0.75\times {10}^{-6}\mathrm{C}}{2\times 16\times {10}^{-6}\mathrm{C}} \\ =2.3\times {10}^{12} \end{gathered}$$

Hence, the value of the number of the alpha particle is $2.3\times {10}^{12}$.

(b) Calculation of the number of alpha particles present in the length of the beam:

So now let N' be the number of particle in length L of the beam.

They will all pass through the beam having cross-section at one end in time

$\mathrm{t}=\mathrm{L}/\mathrm{v}$

Where, v is the particle speed.

But the current is the total charge moving per unit time, so the total charge is,

q = 2eN'

Thus, the value of the total current can be calculated using equation (1) as follows:

$\mathrm{i}=\frac{2\mathrm{eN}\text{'}\mathrm{v}}{\mathrm{L}}$

So, we calculate the number of alpha particle, by rearranging the above equation as follows:

$\mathrm{N}\text{'}=\frac{\mathrm{iL}}{2\mathrm{ev}}$ ….. (5)

The kinetic energy of the alpha particles in terms of joule is given by,

$$\begin{gathered} \mathrm{K}=20\times {10}^6\times 1.6\times {10}^{-19} \\ =32\times {10}^{-13}\mathrm{J} \end{gathered}$$

And mass of alpha particle is four times the mass of proton $\left({\mathrm{m}}_{\mathrm{p}}=1.67\times {10}^{-27}\mathrm{kg}\right)$.

Thus, the mass of the alpha particle is,

$$\begin{gathered} \mathrm{m}=4\times 1.67\times {10}^{-27} \\ =6.68\times {10}^{-27}\mathrm{kg} \end{gathered}$$

Now, the speed of the alpha particle can be given using the data in equation (3) as follows:

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{2\mathrm{K}}{\mathrm{m}}} \\ =\sqrt{\frac{2\times 32\times {10}^{-13}}{6.68\times {10}^{-27}}} \\ =\sqrt{9.61\times {10}^{14}} \\ =3.1\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, the number of alpha particles that is present in the length can be given using equation (5) as follows:

$$\begin{gathered} N\text{'}=\frac{0.25\times {10}^{-6}\times 20\times {10}^{-2}}{2\times 16\times {10}^{-19}\times 3.1\times {10}^7} \\ =\frac{5}{9.92}\times {10}^4 \\ =5\times {10}^3 \end{gathered}$$

Hence, the value of the alpha particles is $5\times {10}^3$.

(c) Calculation of the potential difference:

For finding the potential difference for accelerating the alpha particle, we use law of conservation of energy.

Initial kinetic energy,${\mathrm{K}}_{\mathrm{i}}=0$

Final kinetic energy, ${\mathrm{K}}_{\mathrm{f}}=32\times {10}^{-13}\mathrm{J}$

Final potential energy, ${\mathrm{U}}_{\mathrm{f}}=0$

Initial potential energy,

$$\begin{gathered} {\mathrm{U}}_{\mathrm{i}}=\mathrm{qV} \\ =2\mathrm{eV} \end{gathered}$$

Here, V is the electrical potential through which the particles are accelerated.

Using equation (4), the law of energy conservation, you can get that the potential difference required to accelerate the charged particles is given by,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{f}}={\mathrm{U}}_{\mathrm{i}} \\ {\mathrm{K}}_{\mathrm{f}}=2\mathrm{eV} \\ \mathrm{V}=\frac{{\mathrm{K}}_{\mathrm{f}}}{2\mathrm{e}} \\ =\frac{32\times {10}^{-13}}{2\times 16\times {10}^{-19}} \\ =1\times {10}^7\mathrm{V} \end{gathered}$$

Hence, the value of the potential difference is $1\times {10}^7\mathrm{V}$.