An X-ray tube is operated at `50 kV and 20 m A`. The target material of the tube has mass of `1 kg` and specific heat `495 J kg ^(-1) `^(@)C^(-1)`. One perent of applied electric power is converted into X-rays and the remaining energy goes into heating the target. Then,

Options

A. a suitable target metrial must have hight melting temprature.
B. a suitable target metrial must have low thermal conductivity.
C. the average rate of rice of temprature of the target would be `2 ^(@) C s^(-1)`
D. the minimum wavelength of X-rays emitted is about `0.25 xx 10^(-10) m`

Explanation

To solve the problem step-by-step, we will calculate the average rate of rise of temperature of the target in the X-ray tube.

### Step 1: Calculate the total power supplied to the X-ray tube
The power (P) in an X-ray tube can be calculated using the formula:
\[ P = V \times I \]
where:
- \( V \) is the voltage (50 kV = 50,000 V)
- \( I \) is the current (20 mA = 0.020 A)

Substituting the values:
\[ P = 50,000 \, \text{V} \times 0.020 \, \text{A} = 1000 \, \text{W} \]

### Step 2: Calculate the power converted into heat
According to the problem, only 1% of the total power is converted into X-rays, while the remaining 99% goes into heating the target. Therefore, the power converted into heat (P_heat) is:
\[ P_{\text{heat}} = 0.99 \times P \]
\[ P_{\text{heat}} = 0.99 \times 1000 \, \text{W} = 990 \, \text{W} \]

### Step 3: Use the heat power to find the temperature rise
The temperature rise (\( \Delta T \)) can be calculated using the formula:
\[ P_{\text{heat}} = m \cdot c \cdot \Delta T \]
where:
- \( m \) is the mass of the target (1 kg)
- \( c \) is the specific heat capacity (495 J/kg°C)

Rearranging the formula to solve for \( \Delta T \):
\[ \Delta T = \frac{P_{\text{heat}}}{m \cdot c} \]
Substituting the values:
\[ \Delta T = \frac{990 \, \text{W}}{1 \, \text{kg} \times 495 \, \text{J/kg°C}} \]
\[ \Delta T = \frac{990}{495} \]
\[ \Delta T = 2 \, \text{°C} \]

### Conclusion
The average rate of rise of temperature of the target is \( 2 \, \text{°C} \) per second.