With a d.c power supply giving 10mA at a p.d of 50 kV. The power of the emitted x-ray is 5 watt. The efficiency of the x-ray tube is

Options

A. 0.1
B. 0.01
C. 0.05
D. 0.25

Answer

To find the efficiency of the X-ray tube, we can follow these steps:

### Step 1: Calculate the Supplied Power
The supplied power (P_supplied) can be calculated using the formula:
\[ P_{\text{supplied}} = V \times I \]
where:
- \( V \) is the potential difference (in volts)
- \( I \) is the current (in amperes)

Given:
- \( V = 50 \, \text{kV} = 50 \times 10^3 \, \text{V} = 50000 \, \text{V} \)
- \( I = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} = 0.01 \, \text{A} \)

Substituting the values:
\[ P_{\text{supplied}} = 50000 \, \text{V} \times 0.01 \, \text{A} = 500 \, \text{W} \]

### Step 2: Identify the Emitted Power
The emitted power (P_emitted) is given as:
\[ P_{\text{emitted}} = 5 \, \text{W} \]

### Step 3: Calculate the Efficiency
Efficiency (η) is calculated using the formula:
\[ \eta = \frac{P_{\text{emitted}}}{P_{\text{supplied}}} \]

Substituting the values:
\[ \eta = \frac{5 \, \text{W}}{500 \, \text{W}} = \frac{1}{100} = 0.01 \]

### Step 4: Convert Efficiency to Percentage
To express the efficiency as a percentage:
\[ \eta_{\text{percentage}} = \eta \times 100 = 0.01 \times 100 = 1\% \]

### Final Answer
The efficiency of the X-ray tube is **1%**.

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