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Question: What is the kinetic energy for one electron whose wavelength is \( 650nm \) ? Its rest mass is \( 9....

What is the kinetic energy for one electron whose wavelength is 650nm650nm ? Its rest mass is 9.109×10−34kg9.109 \times {10^{ - 34}}kg .

Explanation

Solution

The relationship between the kinetic energy of a particle and its wavelength can be calculated by taking into account the dual nature of the concerned particle. The kinetic energy is directly linked to the momentum which is linked to the wavelength of the particle.

Complete answer:
De-Broglie was the scientist who presented a mathematical model of the dual nature shown by different particles. For a long time scientists debated on the view of particles having characteristic features of matter or electromagnetic radiations that behave as oscillating waves.
De-Broglie defined the wavelength of the wave-like path followed by particles which depends on their mass and speeds. Thus, he found a relation between the wavelength (λ)(\lambda ) and momentum (p)(p) of the particles. The two attributes showed an inverse relation and the constant used to define an absolute relation was Plank’s constant (h)(h) .
The relationship of De-Broglie’s wavelength and momentum can be written as follows:
λ=hp\lambda = \dfrac{h}{p}
Further, the relationship between kinetic energy (K.E)(K.E) and momentum can be used to derive the desired relationship between wavelength and kinetic energy.
p=2mK.Ep = \sqrt {2mK.E}
On inserting this formula in the above equation we get,
λ=h2mK.E\lambda = \dfrac{h}{{\sqrt {2mK.E} }}
The formula can be rewritten as,
K.E=h22mλ2K.E = \dfrac{{{h^2}}}{{2m{\lambda ^2}}}
On inserting the given values of wavelength, Plank’s constant and mass can be written as follows:
K.E=(6.626×10−34Js)22×9.109×10−34kg×(650×10−9m)2K.E = \dfrac{{{{(6.626 \times {{10}^{ - 34}}Js)}^2}}}{{2 \times 9.109 \times {{10}^{ - 34}}kg \times {{(650 \times {{10}^{ - 9}}m)}^2}}}
Thus, the kinetic energy of the particle comes out to be equal to 5.7×10−25J5.7 \times {10^{ - 25}}J

Note:
The wavelength of any particle is inversely proportional to its mass therefore only tiny particles like protons, alpha particles and electrons are capable of showing the dual behavior in a practical sense. Heavier objects have such low wavelengths that makes their wave nature almost insignificant.