An electron beam travelling at a speed of `10^(7)` m/s strikes a metal target and absorbed by it. If the mass of the target is 0.5 g its specific heatis 100 cal /kg`.^(@)C`, find the rate at which its temperature rises. Given that the electron beam curen is 0.048 A.

Answer

To solve the problem, we need to find the rate at which the temperature of the metal target rises when an electron beam strikes it. We will follow these steps:

### Step 1: Calculate the energy delivered by the electron beam per second.

The energy delivered by the electron beam can be calculated using the formula:

\[ \text{Power} (P) = \text{Current} (I) \times \text{Energy per charge} (E) \]

Where:
- The current \( I = 0.048 \, \text{A} \)
- The energy per charge for an electron can be calculated using its kinetic energy:

\[ E = \frac{1}{2} mv^2 \]

Where:
- \( m \) is the mass of an electron \( \approx 9.11 \times 10^{-31} \, \text{kg} \)
- \( v = 10^7 \, \text{m/s} \)

Calculating \( E \):

\[ E = \frac{1}{2} \times 9.11 \times 10^{-31} \times (10^7)^2 \]
\[ E = \frac{1}{2} \times 9.11 \times 10^{-31} \times 10^{14} \]
\[ E \approx 4.555 \times 10^{-17} \, \text{J} \]

Now, we can calculate the power:

\[ P = I \times E = 0.048 \times 4.555 \times 10^{-17} \]
\[ P \approx 2.18 \times 10^{-18} \, \text{W} \]

### Step 2: Convert the power to calories per second.

1 Joule is approximately equal to 0.239 calories. Therefore, we convert the power:

\[ P \approx 2.18 \times 10^{-18} \, \text{W} \times 0.239 \, \text{cal/J} \]
\[ P \approx 5.21 \times 10^{-19} \, \text{cal/s} \]

### Step 3: Calculate the mass of the target in kg.

The mass of the target is given as 0.5 g, which we convert to kg:

\[ m = 0.5 \, \text{g} = 0.5 \times 10^{-3} \, \text{kg} = 5 \times 10^{-4} \, \text{kg} \]

### Step 4: Calculate the temperature rise using the specific heat formula.

The temperature rise can be calculated using the formula:

\[ \Delta T = \frac{Q}{m \cdot c} \]

Where:
- \( Q \) is the heat absorbed per second (which is the power calculated),
- \( m \) is the mass of the target,
- \( c \) is the specific heat capacity.

Given that the specific heat \( c = 100 \, \text{cal/kg} \cdot {}^\circ C \):

Now substituting the values:

\[ \Delta T = \frac{5.21 \times 10^{-19}}{5 \times 10^{-4} \times 100} \]
\[ \Delta T = \frac{5.21 \times 10^{-19}}{5 \times 10^{-2}} \]
\[ \Delta T \approx 1.042 \times 10^{-17} \, {}^\circ C/s \]

### Final Answer

The rate at which the temperature of the metal target rises is approximately:

\[ \Delta T \approx 1.042 \times 10^{-17} \, {}^\circ C/s \]