Question

In a Coolidge tube, the potential difference across the tube is $20kV$, and $10mA$ current flows through the voltage supply. Only $0.5\%$ of the energy carried by the electrons striking the target is converted into X-rays. The X-ray beam carries a power of
A. $0.1\,W$
B. $1\,W$
C. $2\,W$
D. $10\,W$

Answer 1

A Coolidge tube is a tube that is used for the generation of X-rays in which the cathode tube consists of a spiral filament of incandescent tungsten along with the target that serves as anode consisting of massive tungsten. Here, the values of potential difference and the electric current is given. We will use the formula of power that is the product of potential difference and the electric current.

Formula used:
The formula used for calculating the power is given below
$P = VI$
Here, $P$ is the power, $V$ is the potential difference and $I$ is the current through the tube.

Complete step by step answer:
Consider a Coolidge tube that is used for the generation of X-rays. Here, the potential difference across the tube is $20kV$, and $10mA$ current is given.
Therefore, the potential difference, $V = 20kV = 20 \times {10^3}V$
And, the electric current, $I = 10mA = 10 \times {10^{ - 3}}A$
Now, the formula used for calculating the power drawn by Coolidge tube is given below
$P = VI$
$\Rightarrow \,P = \left( {20 \times {{10}^3}V} \right)\left( {10 \times {{10}^{ - 3}}A} \right)$
$\Rightarrow \,P = 200\,W$
Now, it is given in the question that only $0.5\%$ of the energy carried by the electrons that strike the target is converted into X-rays. Therefore, the power generated by the X-rays when $0.5\%$ of the energy is converted into X-ray beam is given below
$P = \dfrac{{0.5}}{{100}} \times 200$
$\therefore \,P = 1\,W$
Therefore, the power of the X-ray beam generated by the Coolidge tube is $1\,W$.

Hence, option B is the correct answer.

Note: The electrons will be emitted when we will heat the Coolidge tube. Here, you must remember to convert the units and also, you must remember to convert the percentage into normal form. Here, we have first calculated power emitted by tube and then the power of the X-rays.