Question

The energy of a photon is equal to the kinetic energy of a proton. The energy of a photon is $E$ . Let ${\lambda _1}$ be the de Broglie wavelength of the proton and ${\lambda _2}$ be the wavelength of the photon. Then $\dfrac{{{\lambda _1}}}{{{\lambda _2}}}$ is proportional to
(A) ${E^0}$
(B) ${E^{1/2}}$
(C) ${E^{ - 1}}$
(D) ${E^{ - 2}}$

Answer 1

For solving this question, we have to convert ${\lambda _1}$ and ${\lambda _2}$ in terms of $E$ . For that we need to use the Planck-Einstein relation and the de-Broglie equation.

Formula used: The formulae used in solve this question are:
$E = \dfrac{{hc}}{\lambda }$ , where $E$ is the energy of a photon of wavelength $\lambda$ , $h$ is the Planck’s constant, and $c$ is the speed of light.
$\lambda = \dfrac{h}{p}$ , where $\lambda$ is the de Broglie wavelength of a particle having momentum $p$

Complete step by step solution:
The energy of the photon is equal to $E$ .
We know that the energy of a photon is given by $E = \dfrac{{hc}}{\lambda }$
According to the question, $\lambda = {\lambda _2}$
$\therefore E = \dfrac{{hc}}{{{\lambda _2}}}$
$\Rightarrow {\lambda _2} = \dfrac{{hc}}{E}$ …………..(i)
Now, the kinetic energy of the proton $K$ is equal to the energy of the photon.
$\therefore K = E$
We know that the momentum $p$ of a particle is given by
$p = \sqrt {2mK}$
So, the momentum of the proton is equal to
$p = \sqrt {2{m_p}E}$ ( ${m_p}$ is the mass of proton)…………..(ii)
Now, the de Broglie equation is given by
$\lambda = \dfrac{h}{p}$
So, the wavelength ${\lambda _1}$ of the proton is
${\lambda _1} = \dfrac{h}{p}$
Substituting $p$ from (ii)
${\lambda _1} = \dfrac{h}{{\sqrt {2{m_p}E} }}$ …………..(iii)
Dividing (iii) by (i), we get
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\dfrac{h}{{\sqrt {2{m_p}E} }}}}{{\dfrac{{hc}}{E}}}$
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{h}{{\sqrt {2{m_p}E} }} \times \dfrac{E}{{hc}}$
On simplifying, we get
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{1}{c}\sqrt {\dfrac{E}{{2{m_p}}}}$
As $c$ and ${m_p}$ are constants, so
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} \propto \sqrt E$
or
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} \propto {E^{1/2}}$
So, the required ratio is proportional to ${E^{1/2}}$
Hence, the correct answer is option B, ${E^{1/2}}$ .

Note:
For finding the momentum of a photon, the following equation can also be used
$E = pc$
Here the symbols have their usual meanings. The momentum calculated from this equation can be put into the de-Broglie equation to find out the value of the wavelength of the photon. But, always prefer to use the Planck-Einstein equation for a photon, as it directly gives the value of its wavelength.