Question

The de-Broglie wavelength of an electron is the same as that of a 50keV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is 0.5MeV
A- 1:50
B- 1:20
C- 20:1
D- 50:1

Answer 1

Hint : As, in the question we need to find the ratio of the the energy of the photon to the kinetic energy of the electron, i.e., $\dfrac{{{E}_{p}}}{{{E}_{e}}}$ . As, in the question, energy of photons is given and energy of electron mass is given. As, Kinetic energy of electron is calculated by ${{E}_{e}}=\dfrac{{{h}^{2}}}{2m{{\lambda }^{2}}}$ , wavelength is also calculate by ${{\lambda }_{\text{p}}}=\dfrac{hc}{E}$ , and the value of hc= 1240 ev nm. Using these all values the ratio can be calculated.

Complete Step By Step Answer:
Given, Energy of X-ray photon is 50 KeV and Energy equivalent of electron mass is 0.5 MeV.
As,
${{\text{E}}_{\text{p}}}=50\text{KeV}=5\times {{10}^{4}}\text{eV}$
${{\text{E}}_{\text{e}}}=\text{m}{{\text{c}}^{\text{2}}}=0.5\text{MeV}=0.5\times {{10}^{5}}\text{eV}$
As, Wavelength of the photon $\left( {{\lambda }_{\text{photon}}}=\dfrac{hc}{E} \right)$ , as, the value of $hc=1242\text{eV}\cdot \text{nm}$ and ${{\text{E}}_{\text{p}}}=5\times {{10}^{4}}\text{eV}$ .
Use these values to find the value of wavelength of photons.
${{\lambda }_{\text{p}}}=\dfrac{hc}{E} \\\ {{\lambda }_{\text{p}}}=\dfrac{1242}{5\times {{10}^{4}}} \\\ {{\lambda }_{\text{p}}}=0.02484\,\text{nm}$
Therefore, the de-Broglie wavelength of an electron $\lambda ={{\lambda }_{\text{p}}}=0.0248\,\text{nm}$ .
The kinetic energy of electron is given by,
${{E}_{e}}=\dfrac{{{h}^{2}}}{2m{{\lambda }^{2}}}$ , multiply and divide by ${{c}^{2}}$
${{E}_{e}}=\dfrac{{{h}^{2}}{{c}^{2}}}{2m{{c}^{2}}{{\lambda }^{2}}}$
As, ${{\text{E}}_{\text{e}}}=\text{m}{{\text{c}}^{\text{2}}}=0.5\times {{10}^{5}}\text{eV}$ and $hc=1242\text{eV}\cdot \text{nm}$ , so use these values in the above formula and calculate the kinetic energy of photon.
{{E}_{{{e}^{-}}}}=\dfrac{{{h}^{2}}{{c}^{2}}}{2m{{c}^{2}}{{\lambda }^{2}}} \\\ =\dfrac{{{\left( 1242 \right)}^{2}}}{2\left( 0.5\times {{10}^{5}} \right){{\left( 0.02484 \right)}^{2}}} \\\ =2.51\times {{10}^{3}}\text{eV} \
So,
$\dfrac{{{E}_{photon}}}{{{E}_{{{e}^{-}}}}}=\dfrac{5\times {{10}^{4}}}{2.51\times {{10}^{3}}}=20$
So, the ratio of the energy of the photon to the kinetic energy of the electron is $20:1$ .
Therefore, option C is correct.

Note :
The wavelength is dependent upon the frequency and the speed of the propagating wave. Frequency is the characteristic of the source which is producing the wave. The SI unit of the wavelength is metre and velocity of the wave is m/s while for the frequency it is to be taken in Hertz (hz) always.
Also, energy can be taken either in Joules or eV depending upon the demand of the question.