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Question: You are given an electron with a de Broglie wavelength of \( \lambda = 76.3nm \) . What is the Kelvi...

You are given an electron with a de Broglie wavelength of λ=76.3nm\lambda = 76.3nm . What is the Kelvin temperature of this electron ?
(A) 1.50
(B) 2.00
(C) 2.50
(D) 3.00

Explanation

Solution

According to wave-particle duality, the De Broglie wavelength is a wavelength manifested in all the objects in quantum mechanics which determines the probability density of finding the object at a given point of the configuration space. The de Broglie wavelength of a particle is inversely proportional to its momentum

Complete answer:
It is said that matter has a dual nature of wave-particles. de Broglie waves, named after the discoverer Louis de Broglie, is the property of a material object that varies in time or space while behaving similarly to waves. It is also called matter-waves. It holds great similarity to the dual nature of light which behaves as particle and wave, which has been proven experimentally.
In the case of electrons going in circles around the nuclei in atoms, the de Broglie waves exist as a closed-loop, such that they can exist only as standing waves, and fit evenly around the loop. Because of this requirement, the electrons in atoms circle the nucleus in particular configurations, or states, which are called stationary orbits.
We know that,
E=p22m=32kTE = \dfrac{{{p^2}}}{{2m}} = \dfrac{3}{2}kT
On cancelling 22 on both the sides and taking mm on the other side, we get,
p2=3mkT{p^2} = 3mkT
p=3mkT........(1)p = \sqrt {3mkT} ........(1)
We also know that de Broglie wavelength is,
λ=hp\lambda = \dfrac{h}{p}
On putting the value of p (from equation 1) in the above equation, we get,
λ=h3mkT\lambda = \dfrac{h}{{\sqrt {3mkT} }}
Squaring both sides,
λ2=h23mkT{\lambda ^2} = \dfrac{{{h^2}}}{{3mkT}}
This can further be written as,
T=h23mkλ2T = \dfrac{{{h^2}}}{{3mk{\lambda ^2}}}
T=(6.626×10−34)2(3)(1.38×10−23)(9.108×10−31)(76.3×10−9)2T = \dfrac{{{{(6.626 \times {{10}^{ - 34}})}^2}}}{{(3)(1.38 \times {{10}^{ - 23}})(9.108 \times {{10}^{ - 31}}){{(76.3 \times {{10}^{ - 9}})}^2}}}
On solving the above equation, we get,
T=2.00KT = 2.00K
So, the final answer is (B) 2.00.

Note:
De Broglie's hypothesis of matter waves postulates that any particle of matter that has linear momentum is also a wave. The wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The speed of the matter wave is the speed of the particle.