Question

A)What is the de Broglie Hypothesis?
(B) Write the formula for the de Broglie wavelength.
C) Calculate de Broglie wavelength associated with an electron accelerated by a potential difference of 100 Volts.
Given mass of the electron $ = 9.1 \times {10^{ - 31}}kg $ , $ h = 6.634 \times {10^{ - 34}}Js $ , $ 1eV = 1.6 \times {10^{ - 19}}J $.

Answer 1

Hint : De Broglie gave a statement about the wave nature of matter where he gave us a relation between the wavelengths of motion of any object with momentum. The electron volt (eV) is the maximum kinetic energy possessed by an electron when accelerated by a potential difference of 1 volts.

Formula used: In this solution we will be using the following formula;
 $ \lambda = \dfrac{h}{p} $ where $ p $ is the momentum of a body, $ h $ is the Planck constant, and $ \lambda $ is the wavelength of the body.
 $ {p^2} = 2mE $ where $ m $ is the mass of the body, and $ E $ is the kinetic energy of the body.

Complete step by step answer
(A) The de Broglie hypothesis, though mathematical in nature, can be said to state that all matter possesses a wave nature or properties which is dependent on the momentum of the matter particle. Basically, de Broglie said that matters whether microscopic, such as atoms, electrons, protons, or macroscopic such as humans, balls, stones, etc, are associated with waves and thus have wavelength and frequency.
(B) From de Broglie hypothesis, we can derive that the wavelength of a matter particle is given as
 $ \lambda = \dfrac{h}{p} $ where $ p $ is the momentum of a body, $ h $ is the Planck constant, and $ \lambda $ is the wavelength of the body.
(C) An electron is said to be accelerated by a potential difference of 100 volts and we are to calculate the de Broglie wavelength. To do this, we need to use the energy possessed by the electron to calculate the momentum of the particle, then we can use the momentum to calculate the wavelength, as follows:
An electron volt is kinetic energy possessed by an electron when accelerated by a potential of 1 volts. Now, since the electron was accelerated by 100 volts, then the energy is 100 eV. Hence
 $ {E_k} = 100eV = 100 \times 1.6 \times {10^{ - 19}}J $ (since $ 1eV = 1.6 \times {10^{ - 19}}J $ )
Hence, $ {E_k} = 1.6 \times {10^{ - 17}}J $
Momentum is related to kinetic energy as
 $ {p^2} = 2mE $
Then, by inserting all known values, we have
 $ {p^2} = 2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 17}}} \right) = 2.912 \times {10^{ - 47}} $
 $ \Rightarrow p = \sqrt {2.912 \times {{10}^{ - 47}}} = 5.40 \times {10^{ - 24}}kgm/s $
Hence, the wavelength is
 $ \lambda = \dfrac{h}{p} = \dfrac{{6.634 \times {{10}^{ - 34}}}}{{5.40 \times {{10}^{ - 24}}}} = 1.23 \times {10^{ - 10}}m $.

Note
For clarity, in de Broglie’s hypothesis, although we said that all matter possess wavelengths, including macroscopic matter, the wavelength of macroscopic matter is often too small, even at the most minuscule velocity, to be observed.