Q81P - Fundamentals Of Physics

Question

A beam of 16 MeV deuterons from a cyclotron strikes a copper block. The beam is equivalent to current of 15 mA . (a) At what rate do deuterons strike the block? (b) At what rate is thermal energy produced in the block?

Step-by-Step Solution

Answer

1. The rate at which deuterons strike the block is $9.4\times {10}^{13}{\mathrm{s}}^{-1}$
2. The rate at which thermal energy is produced by the block is. 240 W

Step 1: The given data

1. Voltage at which deuteron strikes the block, $\mathrm{V}=16\mathrm{MeV} \mathrm{or} 16\times {10}^6\mathrm{eV}$
2. The deuteron beam is equivalent to the current, $\mathrm{I}=15\mathrm{\mu A} \mathrm{or} 15\times {10}^{-6}\mathrm{A}$

Step 2: Understanding the concept of the thermal energy

By using the relation between current and charge, we can find time, and the inverse of time can give the rate of the strike of deuterons. By using the formula of power, we can find the rate of production of thermal energy.

Formulae:

The current flowing through the area, $I=\frac{q}{t}$                                                 (i)

The electric power due to the change in difference, P = V I                       (ii)

Step 3: a) Calculation of the rate at which the deuterons strike

Charge on deuteron is $1\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$

The rate at which the deuterons strike can be given using the given data in equation (i) as follows:

$$\begin{gathered} \mathrm{t}=\frac{1.6\times {10}^{-19}}{15\times {10}^{-6}} \\ \frac{1}{\mathrm{t}}=\frac{15\times {10}^{-6}}{1.6\times {10}^{-19}} \\ =9.375\times {10}^{13}{\mathrm{s}}^{-1} \\ \approx 9.4\times {10}^{13}{\mathrm{s}}^{-1} \end{gathered}$$

The rate at which deuterons strike the block is $9.4\times {10}^{13}{\mathrm{s}}^{-1}$ .

Step 4: b) Calculation of the rate at which the thermal energy is produced

The rate at which the thermal energy is produced is nothing but the power of the deuterons.

Using all the given values, the rate of the energy can be given using equation (ii) as follows:

$$\begin{gathered} \mathrm{P}=16\times {10}^6\times 15\times {10}^{-6} \\ =240 \mathrm{W} \end{gathered}$$

The rate at which thermal energy is produced by the block is 240W.