Question

The energy of a photon is equal to the kinetic energy of a proton. Let ${\lambda _1}$ be the de- Broglie wavelength of the photon and ${\lambda _2}$ be the wavelength of the radiation and the de-Broglie wavelength of a photon of that radiation?

Answer 1

Hint : We are going to find the de-Broglie wavelength of the photon from the energy relation, then the de – Broglie wavelength of that radiation is found from the energy relation, after that a ratio is calculated for both the wavelengths which gives the de-Broglie wavelength of photon of that radiation.
The de-Broglie wavelength is given by
$\lambda = \dfrac{h}{p}$
The photon energy is given by
$E = h\nu$

Complete Step By Step Answer:
To solve this question, let us start by finding the value of the de-Broglie wavelength of the photon.
The de-Broglie wavelength is given by
$\lambda = \dfrac{h}{p}$
Now, the photon energy is given by
$E = h\nu$
So, the de-Broglie wavelength of photon is
{\lambda _1} = \dfrac{h}{p} \\\ \Rightarrow {\lambda _1} = \dfrac{h}{{\sqrt {2mE} }} - - - (1) \\\
Now the de-Broglie wavelength of that radiation is
$E = h{\nu _2}$
Simplifying this relation in order to get the wavelength
E = \dfrac{{hc}}{{{\lambda _2}}} \\\ \Rightarrow {\lambda _2} = \dfrac{{hc}}{E} - - - (2) \\\
Now, if we divide the equations $(1)$ and $(2)$ , we get
\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{h}{{\sqrt {2mE} }} \times \dfrac{E}{{hc}} \\\ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} \sim \dfrac{E}{{\sqrt E }} \\\ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt E \\\
Hence, the de-Broglie wavelength of the photon of that radiation will be equal to $\sqrt E$ .

Note :
The de-Broglie wavelength of a radiation can be found if the energy or momentum of the particles are given. If we know the de- Broglie wavelength of a photon and the de-Broglie wavelength of a radiation, then, in order to find the de –Broglie wavelength of a photon of that radiation, we divide the de-Broglie wavelength of a photon with the de-Broglie wavelength of the radiation.