Question

What is the de Broglie hypothesis?
(A) Write the formula for the de Broglie wavelength.
(B) Calculate de Broglie wavelength associated with the electron accelerated by a potential difference of 100 volts.
(C) Given Mass of the electrons= 9.1×10$^{-31}$ kg, h=6.634×10$^{-34}$ Js, 1eV=1.6×10$^{-19}$

Answer 1

Hint: The dual nature of the matter wave is explained by de Broglie’s hypothesis. The wavelength of the matter wave is known as the de-Broglie’s wavelength. The expression for de-Broglie’s wavelength is given as:
$\lambda =\dfrac{h}{p}$

Complete step-by-step answer:
A.
The de Broglie hypothesis states that the matter can have dual nature. The matter can behave as particles and as waves in quantum mechanics.
Consider the table have a plank fixed at both ends. The quantum level or the microscopic level observation regarding this is true. But the macroscopic observation of this becomes very difficult and complicated.
It implies that the matter can have dual nature as the particle and as the wave.

B.
The de Broglie’s wavelength is given as:
$\Rightarrow \lambda =\dfrac{h}{p}$
Where, h is the planck's constant and p is the momentum

C.
Given:
Voltage, V = 100 volt
Mass of the electrons= 9.1×10$^{-31}$ kg,
Planck's constant, h=6.634×10$^{-34}$ Js,
1eV=1.6×10$^{-19}$
Calculation:
The de Broglie’s wavelength is given as:
$\begin{align}
  & \Rightarrow \lambda =\dfrac{h}{p} \\
 & \Rightarrow \lambda =\dfrac{h}{\sqrt{2mE}} \\
 & \Rightarrow \lambda =\dfrac{h}{\sqrt{2mqV}} \\
 & \Rightarrow \lambda =\dfrac{6.63\times {{10}^{-34}}\ \text{Js}}{\sqrt{2\times 9.1\times {{10}^{-31}}\ \text{kg}\times \text{50}\times \text{1}\text{.6}\times \text{1}{{\text{0}}^{-19}}}} \\
 & \Rightarrow \lambda =1.72\ \text{Angstrom} \\
\end{align}$

Note:
Another method:
The momentum of the electron is given as:
$\begin{align}
  & \Rightarrow p=\sqrt{2mE} \\
 & \Rightarrow p=\sqrt{2\times 9.1\times {{10}^{-31}}\ \text{kg}\times \text{50}\times \text{1}\text{.6}\times \text{1}{{\text{0}}^{-19}}} \\
 & \Rightarrow p=38.15\times {{10}^{-20}}\ \text{kgm}{{\text{s}}^{\text{-1}}} \\
\end{align}$

The de Broglie’s wavelength is given as:
\[\begin{align}
  & \Rightarrow \lambda =\dfrac{h}{p} \\
 & \Rightarrow \lambda =1.72\ \text{Angstrom} \\
\end{align}\]