Question

An x-ray tube with a tungsten target (Z=74, density =19.25 g/cm³, specific heat =0.134 kJ/kgK) is operated at 100kV and a tube current of 100 mA for 2 s. If the tube voltage is dropped to 50 kV and the tube current is increased to 200 mA, the exposure time required to produce the same number of Bremsstrahlung photons is most nearly (a) 0.5 s (b) 1.0 s (c) 2.0 s (d) 4.0 s (e) 8.0 s

Answer 1

The number of Bremsstrahlung photons produced in an x-ray tube depends on the energy deposited in the target, which is the product of the voltage (V), current (I), and time (t).

Given that the initial conditions are V1 = 100 kV (kilo volts), I1 = 100 mA (milli amperes), and t1 = 2 s (seconds) and that the final conditions are V2 = 50 kV, I2 = 200 mA, with the exposure time, t2, unknown.

We know that the energy deposited to produce the same number of Bremsstrahlung photons in both conditions has to be equal. Therefore, we can set up the equation V1*I1*t1 = V2*I2*t2.

To find t2, we rearrange our equation to isolate it on one side: t2 = (V1*I1*t1) / (V2*I2).

Now we can substitute in our given values and calculate:

Substituting,
t2 = (100 kV * 100 mA * 2 s) / (50 kV * 200 mA)

Which simplifies to:

t2 = 2s

So, the time required to produce the same number of Bremsstrahlung photons when the tube voltage is dropped to 50 kV and current is increased to 200 mA is 2 seconds. Hence, the answer is (c) 2.0s.