Question

Calculate the
(a) Momentum
(b) De Broglie wavelength of the electronics accelerated through a potential difference of \[56V\].

Answer 1

Hint: Momentum is a vector quantity, it is defined as the product of the mass of an object and its velocity. The body’s momentum is always in the same direction as its vector velocity. It’s a conserved quantity, as the total momentum of a system is constant.
De Broglie waves exist as a closed-loop where the electrons are going in circles around the nuclei in atoms. The wave can exist only as standing waves fitting evenly around the loop. Because of this requirement, the electrons in atoms that circle the nucleus are in stationary orbits.
Formula used:
The formula of momentum is
\[P = mv\]
The de Broglie wavelength formula is expressed as
\[\lambda = \dfrac{h}{P}\]

Complete answer:
(a)
At the equilibrium, kinetic energy is accelerating potential.
\[\dfrac{{m{v^2}}}{2} = eV\]
As we know the potential difference, \[V = 56V\], the mass of the electron \[m = 9.109 \times {10^{ - 31}}kg\]
\[ \Rightarrow v = 4.44 \times {10^6}m/s\]
The momentum expresses with the letter \[P\] and is used as momentum for short. And the mass is represented with the letter \[m\] and velocity \[v\].
Now, applying the formula of momentum we get,
\[P = mv\]
\[P = 9.109 \times {10^{ - 31}} \times 4.44 \times {10^6}kg.m/s\]
\[P = 4.04kg.m/s\]
Hence, the calculated momentum is \[P = 4.04kg.m/s\]
(b)
According to de Broglie, like light, matter also can show wave-particle duality. As we know light can behave both as a wave and as a particle. That's the reason that matter follows the same equation for wavelength as light.
The Plank constant \[h = 6.626070 \times {10^{ - 34}}J/Hz\]. And the momentum of the electron as we calculate previously \[P = 4.04 \times {10^{ - 24}}kg.m/s\]
From the equation of wavelength we get,
\[\lambda = \dfrac{h}{P}\]
\[\lambda = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{4.04 \times {{10}^{ - 24}}}}m\]
\[\lambda = 1.64 \times {10^{ - 10}}m\]
\[\lambda = 0.164nm\]
The calculated wavelength is \[0.164nm\].

Note:
The momentum of an object has a directly proportional relationship to its mass and velocity. Thus the greater an object’s mass or velocity, the greater its momentum. \[kg.m/s\] is the S.I unit of momentum.
De Broglie's wavelength is important in quantum mechanics. The wavelength \[\left( \lambda \right)\] of an object is in relation to its mass and momentum, which is known as de Broglie wavelength. A particle’s de Broglie wavelength is inversely proportional to its force.