Question

An electron and a photon, each has a de-Broglie wavelength of $1.2A°$. The ratio of their energies will be:
A) $1:1$
B) $1:10$
C) $1:100$
D) $1:1000$

Answer 1

The de-Broglie equation states that a matter can act as waves as well as particles same as that of light and radiation. Every moving particle has a wavelength. This equation is one of the important equations that is used to define the properties of matter.

Complete step by step solution:
The particles that have very low mass move at a less speed than the speed of light. The De-Broglie equation gives a relationship between the mass and the wavelength of the particle. The wavelength for a photon is given by
$\lambda = \dfrac{h}{p}$
Or $\lambda = \dfrac{h}{{\dfrac{{{E_p}}}{c}}} = \dfrac{{hc}}{{{E_p}}}$---(i)
Where ‘h’ is Planck’s constant
‘c’ is the speed of light
${E_p}$is the energy of proton
The De-Broglie wavelength for an electron is given by
$\lambda = \dfrac{h}{p}$
‘h’ is Planck’s constant
‘p’ is the momentum
Or $\lambda = \dfrac{h}{{\sqrt {2m{E_e}} }}$---(ii)
Squaring equation (ii), it becomes
${\lambda ^2} = \dfrac{{{h^2}}}{{2m{E_e}}}$---(iii)
Dividing equation (i) and equation (iii) and calculating their ratios,
$\Rightarrow \dfrac{\lambda }{{{\lambda ^2}}} = \dfrac{{\dfrac{{hc}}{{{E_p}}}}}{{\dfrac{{{h^2}}}{{2m{E_e}}}}}$
$\Rightarrow \dfrac{1}{\lambda } = \dfrac{{hc}}{{{E_p}}} \times \dfrac{{2m{E_e}}}{{{h^2}}}$
$\Rightarrow \dfrac{1}{\lambda } = 2mc\dfrac{{{E_e}}}{{h{E_p}}}$
$\Rightarrow \dfrac{{{E_e}}}{{{E_p}}} = \dfrac{h}{{2mc\lambda }}$---(iv)
Given that the de-Broglie wavelength is$\lambda = 1.2A = 1.2 \times {10^{ - 10}}m$
Mass of the electron is $m = 9.1 \times {10^{ - 31}}kg$
Speed of the light is $c = 3 \times {10^8}m/s$
$h = 6.62 \times {10^{ - 34}}Js$
Substituting all the values in equation (iv) and solving for the ratio,
$\Rightarrow \dfrac{{{E_e}}}{{{E_p}}} = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{2 \times 9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^8} \times 1.2 \times {{10}^{ - 10}}}}$
$\Rightarrow \dfrac{{{E_e}}}{{{E_p}}} = \dfrac{1}{{100}}$
Or ${E_e}:{E_p} = 1:100$

Therefore, Option (C) is the right answer.

Note: It is to be noted that the electrons and photons are microscopic particles. They possess a dual nature property. This means that they have wavelength and also have frequency. Any particle that is moving will have a wave character and is called matter waves. The wave and the particle nature of the matter are complementary to each other.