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Question: An X-ray tube operated at \(50kV\), produces heat at the target at the rate of \(796W\). If \(0.5\% ...

An X-ray tube operated at 50kV50kV, produces heat at the target at the rate of 796W796W. If 0.5%0.5\% of the energy of electrons is used for X-ray generation, what is the number of electrons striking the target per second?
a. 1019 {10^{19}}
b. 1018{10^{18}}
c. 1017{10^{17}}
d. 1016{10^{16}}

Explanation

Solution

Hint: Only some part of the total energy supplied is generated out as heat(0.5%)(0.5\% ) . The rest of the energy(99.5%)(99.5\% ) will be passed on to the electrons which will hit the target surface to produce the X-rays. Also, the stream of moving electrons can also be considered as current flowing.

Complete Step by Step Solution:
X-rays are produced due to sudden deceleration of fast-moving electrons when they collide and interact with the target anode. The major part of the energy is generated out as heat and the rest part is used for generating X-rays.

It is given that the heat generated(H)(H) is equal to 796W796W and 0.5%0.5\% of the total energy is generated out as heat.

Therefore, it can be rightly said that the remaining part of the total energy is generated as heat.
H=99.5%H = 99.5\% of the total energy =796W = 796W per second

Now, the total energy that is supplied per second is equal to the power supplied to the system.
Power supplied(P)=VI(P) = VI, where VV is the potential at which the X-ray tube is operated and II is the current flowing.

It is given that V=50kV=50×103VV = 50kV = 50 \times {10^3}V
Since the current that is flowing actually consists of electrons as moving charges, the current(I)(I) can be expressed as a collection of several electrons. I=neI = ne , where nn is the number of electrons striking the target per second and ee is the charge of one electron (e=1.6×10−19C)(e = 1.6 \times {10^{ - 19}}C).

So, P=V×n×eP = V \times n \times e

We can say that the 99.5%99.5\% of the total energy supplied per second is equal to 796W796W.
99.5%99.5\% of P=796WP = 796W
99.5100×V×n×e=796\dfrac{{99.5}}{{100}} \times V \times n \times e = 796
99.5100×50×103×n×1.6×10−19=796\dfrac{{99.5}}{{100}} \times 50 \times {10^3} \times n \times 1.6 \times {10^{ - 19}} = 796
n=796×10099.5×50×103×1.6×10−19n = \dfrac{{796 \times 100}}{{99.5 \times 50 \times {{10}^3} \times 1.6 \times {{10}^{ - 19}}}}
n=10×1016n = 10 \times {10^{16}}
n=1017n = {10^{17}}
So, Option (C)(C) is correct.

Note: The concepts used in the process of X-rays production are essential for problems like these. The student should carefully note that only some part of the total energy supplied is passed on to the electrons. Also, the relation between power and energy supplied may confuse the student, but a simple definition regarding this will help him/her to avoid any confusion.
The stream of moving electrons can be considered as current flowing and vice versa, so that the equations of power supplied can be applied in order to approach the problem.