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

Hint: 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.