Question

The de-Broglie wavelength of an electron moving with a velocity of $1.5\times {{10}^{8}}m/s$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon $c=3\times {{10}^{8}}m/s$:
(a) $2$
(b) $4$
(c) $\dfrac{1}{2}$
(d) $\dfrac{1}{4}$

Answer 1

Hint: First of all we will find the value of wavelength of electron by using the formula, $\lambda =\dfrac{h}{p}=\dfrac{c}{v}$, then we will compare the wavelength of electron and proton as it is give in the question and then we will find the kinetic energy of proton and electron by using the formula, $k=\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}p{{v}_{e}}$ and then we will take their ratios we will find the answer.

Formula used:
$\lambda =\dfrac{h}{p}=\dfrac{c}{v}$, $k=\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}p{{v}_{e}}$

Complete step by step answer:
In the question we are given that velocity of a de-Broglie wavelength of an electron is $1.5\times {{10}^{8}}m/s$ and we are also given that wavelength of an electron is equal to proton so, first of all we will find the values of wavelength of electron and proton by using the formula,
$\lambda =\dfrac{h}{p}=\dfrac{c}{v}\Rightarrow p=\dfrac{h}{\lambda }$ ………………………(i)
Where, $\lambda $ is wavelength, h is planck's constant, p is momentum, c is velocity of light, and v is velocity of particle,
And ${{\lambda }_{p}}={{\lambda }_{e}}$
Now, the kinetic energy of electron and proton can be given by the formula,
$k=\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}p{{v}_{e}}$ …………………………….(ii)
Now, on substituting the value of p from expression (i) in expression (ii) we will get the kinetic energy of electron as,
\[K.E{{.}_{e}}=\dfrac{h{{v}_{e}}}{2{{\lambda }_{e}}}\Rightarrow {{\lambda }_{e}}=\dfrac{h{{v}_{e}}}{2K.E{{.}_{e}}}\]
And kinetic energy of proton can be given as,
\[K.E{{.}_{p}}=\dfrac{hc}{2{{\lambda }_{p}}}\Rightarrow {{\lambda }_{e}}=\dfrac{hc}{K.E{{.}_{p}}}\]
Now, on comparing the values of wavelength as they are equal, we will get,
\[\dfrac{hc}{K.E{{.}_{p}}}=\dfrac{h{{v}_{e}}}{2K.E{{.}_{e}}}\]
\[\Rightarrow \dfrac{2hc}{h{{v}_{e}}}=\dfrac{K.E{{.}_{p}}}{K.E{{.}_{e}}}\Rightarrow \dfrac{K.E{{.}_{p}}}{K.E{{.}_{e}}}=\dfrac{2c}{{{v}_{e}}}\]
Now, on substituting the values of c and v we will get,
\[\Rightarrow \dfrac{K.E{{.}_{p}}}{K.E{{.}_{e}}}=\dfrac{2\times 3\times {{10}^{8}}}{1.5\times {{10}^{6}}}=4\]
Hence, the ratio of kinetic energy of photon and electron is 4.
Thus, option (b) is the correct answer.

Note: Wave energy of photon in de-Broglie’s wavelength can be given by the formula, in which v is velocity of electron or photon. Now in the case of photons, according to de-Broglie, light is considered as packets of energy which travel in the form of waves with a specific wavelength. This nature of light is different from the normal nature of light in the form of rays.