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Question: A beam of helium atoms moves with a velocity of \[2 \times {10^4}\]. Find the wavelength of particle...

A beam of helium atoms moves with a velocity of 2×1042 \times {10^4}. Find the wavelength of particles constituting with the beam.

Explanation

Solution

We are given the velocity of a beam of helium atoms. With the help of the given data we have to determine the wavelength of the particles constituting the beam. We know that we can determine the wavelength of moving particles with the help of the de Broglie wave equation.

Complete step-by-step solution: De Broglie equation gives a relationship between the momentum and wavelength of a moving particle. Actually it helps us to know that matter have a wavelength explaining that in the same manner as the beam of light gets diffracted, the beam of electrons can also be diffracted. In short, we can conclude that it tells us that ever moving particle has a wavelength. Thus, wave properties of a matter or say matter waves can be described with the help of de Broglie equation.
Therefore, the de Broglie wavelength of particles constituting the beam can be derived with the help of de Broglie equation which is mathematically expressed as;
λ=h/mv\lambda = h/mv
Where, λ\lambda is de Broglie wavelength.
h is Planck’s constant (6.626×10−34Js)\left( {6.626 \times {{10}^{ - 34}}J{\rm{ }}s{\rm{ }}} \right).
m is the mass of the particle.
and v is the velocity of the particle.
We know that the mass of a Helium atom is 4 amu. The conversion of amu to kg is done as shown below.:
4  amu×1  g6.022×1023  amu =0.664×10−23g =6.64×10−27kg 4\;amu \times \dfrac{{1\;g}}{{6.022 \times {{10}^{23}}\;amu}}\\\ = 0.664 \times {10^{ - 23}}g\\\ = 6.64 \times {10^{ - 27}}kg
Putting the above values in the de Broglie equation we have;
λ=hm×v ⇒λ=6.626×10−34  kg  m2  s−16.64×10−27  kg×2×104  m/s ⇒λ=6.626×10−34  kg  m2  s−11.328×10−22  kg  m/s ∴λ=4.98×10−12  m \lambda = \dfrac{h}{{m \times v}}\\\ \Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}\;kg\;{m^2}\;{s^{ - 1}}}}{{6.64 \times {{10}^{ - 27}}\;kg \times 2 \times {{10}^4}\;m/s}}\\\ \Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}\;kg\;{m^2}\;{s^{ - 1}}}}{{1.328 \times {{10}^{ - 22}}\;kg\;m/s}}\\\ \therefore \lambda = 4.98 \times {10^{ - 12}}\;m

Hence, the wavelength of the particles constituting the beam is4.98×10−12  m4.98 \times {10^{ - 12}}\;m.

Note: This de-Broglie wavelength provides the information that photons must have wave nature in addition to particle nature and also have a wavelength known as de-Broglie wavelength.