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Question: The de-Broglie wavelength of a particle of mass 6.63 g moving with a velocity of 100 m/s is: \( ...

The de-Broglie wavelength of a particle of mass 6.63 g moving with a velocity of 100 m/s is:
A. 10−33m B. 10−35m C. 10−31m D. 10−25m  {\text{A}}{\text{. 1}}{{\text{0}}^{ - 33}}m \\\ {\text{B}}{\text{. 1}}{{\text{0}}^{ - 35}}m \\\ {\text{C}}{\text{. 1}}{{\text{0}}^{ - 31}}m \\\ {\text{D}}{\text{. 1}}{{\text{0}}^{ - 25}}m \\\

Explanation

Solution

The De-Broglie wavelength of a particle is equal to the Planck’s constant divided by the momentum of the particle. The momentum of a particle is equal to the product of its mass and the velocity. Using given information in this formula, we can obtain the required solution.

Formula used:
The de-Broglie wavelength of particle of mass m and having velocity v is given as
λ=hp=hmv\lambda = \dfrac{h}{p} = \dfrac{h}{{mv}}

Complete answer:
We are given a particle whose mass is given as
m=6.63g=6.63×10−3kgm = 6.63g = 6.63 \times {10^{ - 3}}kg
This particle is travelling with a velocity v whose value is given as
v=100m/sv = 100m/s
We know the formula for the de-Broglie wavelength of a particle which is given as
λ=hp=hmv\lambda = \dfrac{h}{p} = \dfrac{h}{{mv}}
Here p represents the momentum of the given particle which is equal to the product of the mass and the velocity of the given particle; h is known as the Planck’s constant whose value is given as
h=6.626×10−34Jsh = 6.626 \times {10^{ - 34}}Js
Now we can directly substitute these values into the formula for the de-Broglie wavelength of the given particle. Doing so, we get
λ=6.63×10−346.63×10−3×100=10−33m\lambda = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{6.63 \times {{10}^{ - 3}} \times 100}} = {10^{ - 33}}m
This is the required value of the de-Broglie wavelength of the given particle.

Hence, the correct answer is option A.

Additional information:
De-Broglie postulated that with every matter particle, there is a wavelength associated called the de-Broglie wavelength. It plays a significant role when we measure it for subatomic particles in the quantum regime.

Note:
1. All units should be in the same system of units.
2. It should be noted that the de-Broglie wavelength obtained for the given particle is very small. It depends on the mass of the particle and its velocity. Larger the values of the mass and/or the velocity of the given particle, smaller is the obtained de-Broglie wavelength.