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Question: An X-ray tube operating at \[20\,{\text{kV}}\] shows a current \[1\,{\text{mA}}\]. Assuming efficien...

An X-ray tube operating at 20kV20\,{\text{kV}} shows a current 1mA1\,{\text{mA}}. Assuming efficiency 1%, find the number of X-ray photons emitted per second.
A. 5.25×10135.25 \times {10^{13}}
B. 6.25×10136.25 \times {10^{13}}
C. 6.25×10156.25 \times {10^{15}}
D. 5.25×10155.25 \times {10^{15}}

Explanation

Solution

Use the formula for the electric current. This formula gives the relation between the electric current, number of electrons and charge on one electron. Using this formula, calculate the number of electrons corresponding to the given value of the electric current. One electron corresponds to one photon. Hence, using the value of the efficiency of the X-ray tube, calculate the value of the number of photons emitted per second.

Formula used:
The current II in given by
I=neI = ne
Here, nn is the number of electrons and ee is the charge on one electron.

Complete step by step answer:
We have given that the operating voltage of the X-ray tube is 20kV20\,{\text{kV}}.
V=20kVV = 20\,{\text{kV}}
The current shown by the X-ray tube is 1mA1\,{\text{mA}}.
I=1mAI = 1\,{\text{mA}}
The efficiency of the X-ray tube is 1%.Let us first calculate the number nn of electrons corresponding to the electric current shown by the X-ray tube.Convert the unit of the current shown by the X-ray tube in the SI system of units.
I=(1mA)(103A1mA)I = \left( {1\,{\text{mA}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{A}}}}{{1\,{\text{mA}}}}} \right)
I=103A\Rightarrow I = {10^{ - 3}}\,{\text{A}}
Hence, the current shown by the X-ray tube is 103A{10^{ - 3}}\,{\text{A}}.
Rearrange equation (1) for the number of electrons.
n=Ien = \dfrac{I}{e}
Substitute 103A{10^{ - 3}}\,{\text{A}} for II in the above equation.
n=103Aen = \dfrac{{{{10}^{ - 3}}\,{\text{A}}}}{e}
n=11000e\Rightarrow n = \dfrac{1}{{1000e}}
Hence, the number of electrons are 11000e\dfrac{1}{{1000e}}.

We have given that the efficiency of the X-ray tube is 1%.
η=0.01\eta = 0.01
In an ideal condition, we consider that each electron in the X-ray tube corresponds to a photon. But in this case, the efficiency of the X-ray tube is 1%. Hence, the number NN of photons emitted are
N=11000e×ηN = \dfrac{1}{{1000e}} \times \eta
N=11000e×0.01\Rightarrow N = \dfrac{1}{{1000e}} \times 0.01
Substitute 1.6×1019C1.6 \times {10^{ - 19}}\,{\text{C}} for ee in the above equation.
N=11000(1.6×1019C)×0.01\Rightarrow N = \dfrac{1}{{1000\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)}} \times 0.01
N=6.25×1013\therefore N = 6.25 \times {10^{13}}
Therefore, the number of photons emitted per second are 6.25×10136.25 \times {10^{13}}.

Hence, the correct option is B.

Note: The students may get confused and assume that the number of electrons we have calculated are the number of photons emitted per second. But the students should keep in mind that the number of electrons we have calculated are corresponding to the electric current. But the number of photons emitted depends on the number of electrons incident on the metal plate from which the photons are emitted to form an X-ray.