An X-ray tube when operated at 60 kV, then reading of tube current is 50 mA. Assuming that the total energy of electron is converted into heat, the rate of heat produced at the anode in calories/second is about :

Options

A. 100
B. 300
C. 500
D. 714

Answer

To solve the problem, we need to calculate the rate of heat produced at the anode of the X-ray tube when it is operated at 60 kV with a tube current of 50 mA. We will follow these steps:

### Step 1: Understand the relationship between energy, charge, and potential difference
The energy (E) converted to heat in the X-ray tube can be calculated using the formula:
\[ E = q \times V \]
where:
- \( E \) is the energy in joules,
- \( q \) is the charge in coulombs,
- \( V \) is the potential difference in volts.

### Step 2: Calculate the charge (q)
The charge can be calculated from the tube current (I) using the formula:
\[ I = \frac{q}{t} \]
For a current of 50 mA (which is 0.050 A), and considering a time period of 1 second:
\[ q = I \times t = 0.050 \, \text{A} \times 1 \, \text{s} = 0.050 \, \text{C} \]

### Step 3: Convert the potential difference (V) to volts
The potential difference is given as 60 kV, which we convert to volts:
\[ V = 60 \, \text{kV} = 60 \times 10^3 \, \text{V} = 60000 \, \text{V} \]

### Step 4: Calculate the energy (E) in joules
Now we can substitute the values of \( q \) and \( V \) into the energy formula:
\[ E = q \times V = 0.050 \, \text{C} \times 60000 \, \text{V} \]
\[ E = 3000 \, \text{J} \]

### Step 5: Convert energy from joules to calories
To convert joules to calories, we use the conversion factor:
\[ 1 \, \text{calorie} = 4.2 \, \text{joules} \]
Thus, to convert 3000 joules to calories:
\[ \text{Energy in calories} = \frac{3000 \, \text{J}}{4.2 \, \text{J/cal}} \]
\[ \text{Energy in calories} \approx 714.29 \, \text{cal} \]

### Conclusion
The rate of heat produced at the anode in calories per second is approximately **714 calories/second**.

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