Chapter 12, Problem 12.1QAP

Interpretation Introduction

Interpretation:

If the X-ray with tungsten target is operating at a voltage of 50 kV, the short wavelength limit of the produced continuum needs to be calculated.

Concept introduction:

Short wavelength limit is calculated by using Planck’s constant using the following formula-

${\lambda }_{SWL}=\frac{12.398}{V}$

Here,

${\lambda }_{SWL}$ = short wavelength limit

V= voltage

Answer to Problem 12.1QAP

Short wavelength limit of continuum is $\text{0}\text{.248}\stackrel{\text{o}}{\text{A}}$.

Explanation of Solution

Short wavelength in X-ray tube is defined as the condition at which the intensity of spectrum is zero at a certain wavelength, where electrons transfer their energy into photon energy. This is given as-

$eV=h{v}_{\max }$

On putting the values,

${\lambda }_{SWL}=\frac{12.398}{V}$ …… (1)

$V=5{\text{0 kV= 5×10}}^{\text{3}}\text{ V}$

From equation (1)

${\lambda }_{SWL}=\frac{12.398}{50\times {10}^3}$

Therefore,

${\lambda }_{SWL}=\text{0}\text{.248}\stackrel{\text{o}}{\text{A}}$

Conclusion

Short wavelength of continuum is calculated by using Planck’s constant formula which is the ratio of constant value and the given voltage value. Therefore, calculated value of short wavelength is ${\lambda }_{SWL}=0.\text{248}\stackrel{\text{o}}{\text{A}}$.