Chapter 27, Problem 78GP

(a)

To determine

To show: The photon of shortest wavelength will have: ${\lambda }_0=\frac{hc}{eV}$

Explanation of Solution

Given:

In an X-ray tube, the high voltage is $V$

Formula used:

Total energy of a photon is given by: $E=hf$

Here,

  $h$ is Planck’s constant

  $f$ is frequency

Calculation:

The maximum energy of an electron can be given as: $E=eV$

Equating this expression to the above formula,

  $eV=hf$

Now, frequency can be written as: $f=\frac{c}{{\lambda }_0}$

So, the wavelength can be given as:

  $\begin{array}{l}eV=h\left(\frac{c}{{\lambda }_0}\right)\\ {\lambda }_0=\frac{hc}{eV}\end{array}$

Conclusion:

Thus, the photon of shortest wavelength will have ${\lambda }_0=\frac{hc}{eV}$

(b)

To determine

The shortest wavelength of X-ray emitted.

Answer to Problem 78GP

The shortest wavelength of X-ray emitted is ${\lambda }_0=0.0414\text{ nm}$

Explanation of Solution

Given:

Voltage, $V=30\text{ kV}=30\times {10}^3\text{V}$

Known data:

Speed of light, $c=3\times {10}^8\text{m/s}$

Planck’s constant, $h=6.626\times {10}^{-34}\text{kg}\cdot {\text{m}}^2\text{/s}$

Charge of electron, $e=1.6\times {10}^{-19}\text{C}$

Formula used:

The shortest wavelength is given as: ${\lambda }_0=\frac{hc}{eV}$

Calculation:

Substituting all the given values,

  $\begin{array}{l}{\lambda }_0=\frac{\left(6.626\times {10}^{-34}\text{kg}\cdot {\text{m}}^2\text{/s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(1.6\times {10}^{-19}\text{C}\right)\left(30\times {10}^3\text{V}\right)}\\ {\lambda }_0=0.0414\times {10}^{-9}\text{m}\\ {\lambda }_0=0.0414\text{ nm}\end{array}$

Conclusion:

Thus, the shortest wavelength of X-ray emitted is ${\lambda }_0=0.0414\text{ nm}$