Question

An electron and a photon have the same wavelength. If $p$ is the momentum of electron and $E$ is the energy of the photon, the magnitude of $p/E$ in SI unit is
$A)\text{ }3.0\times {{10}^{8}}$
$B)\text{ }3.33\times {{10}^{-9}}$
$C)\text{ 9}.1\times {{10}^{-31}}$
$D)\text{ 6}.64\times {{10}^{-34}}$

Answer 1

Hint: This problem can be easily solved by using the direct formula for the energy of a photon and the momentum of an electron in terms of their wavelengths. After getting the formulas, we can divide to get the required ratio.

Formula used:
$p=\dfrac{h}{\lambda }$
$E=\dfrac{hc}{\lambda }$
Complete step by step answer:
We will apply the direct formulae for the momentum of a particle and the energy of a photon in terms of their wavelengths.
The momentum $p$ of a particle is given by
$p=\dfrac{h}{\lambda }$ --(1)
where $\lambda$ is the wavelength of the particle and $h=6.636\times {{10}^{-34}}J.s$.
The energy $E$ of a photon is given by
$E=\dfrac{hc}{\lambda }$ --(2)
Where $c=3\times {{10}^{8}}m/s$ and $h=6.636\times {{10}^{-34}}J.s$.
Let the energy of the photon be $E$.
The momentum of the electron be $p$.
Since, it is given that both the electron and photon have the same wavelength.
Let their wavelengths be $\lambda$.
Using (1), we get,
$p=\dfrac{h}{\lambda }$ --(3)
Also, using (2), we get,
$E=\dfrac{hc}{\lambda }$ --(4)
Therefore, using (3) and (4), we get,
$\dfrac{p}{E}=\dfrac{\dfrac{h}{\lambda }}{\dfrac{hc}{\lambda }}=\dfrac{1}{c}=\dfrac{1}{3\times {{10}^{8}}}=0.333\times {{10}^{-8}}=3.33\times {{10}^{-9}}\text{ SI units}$
Hence, the required ratio is $3.33\times {{10}^{-9}}$.
Therefore, the correct option is $B)\text{ }3.33\times {{10}^{-9}}$.

Note: Students must keep in mind that all particles have an inherent wavelength that is related to their momentum inversely. However, this wavelength is not easily observable for daily objects since they have a very large momentum and hence, the wavelength is extremely small and cannot be observed. Students have also got an expression for the energy of a photon. This energy is carried by photons and can be transferred to other objects when radiation is incident on them. This is why we feel warm when sunlight falls on our hands. Sunlight transfers thermal energy to our hands.