Question

Calculate the kinetic energy of $\alpha$ particle which has a wavelength of 12pm

Answer 1

To solve this question, we need to know about the de-Broglie wavelength. It states that a matter can act as waves much like light and radiation which also behave as waves and particles. So, if we look at every moving particle whether it is microscopic or macroscopic, it will have a wavelength.

Formula used:
$\lambda = \dfrac{h}{{\sqrt {2ME} }}$
Further, $E = \dfrac{{{h^2}}}{{2m{\lambda ^2}}}$ where h is Planck’s constant

Complete step by step answer:
In general, the de Broglie equation explains that a beam of electrons can be diffracted just like a beam of light. It helps us to understand the idea of matter having a wavelength. Moreover, Louis de Broglie in his thesis suggested that any moving particle, whether microscopic or macroscopic, will be associated with a wave character known as matter waves. He further proposed a relation between the velocity and momentum of a particle with the wavelength if the particle had to behave as a wave.
Now, we have to find the kinetic energy of the $\alpha$ particle which has a wavelength of 12pm. So according to the formula,
$\lambda = \dfrac{h}{{\sqrt {2ME} }}$
Where, $\lambda = 12pm$
Now, mass of the particle i.e. mass of ${H_e} = 4amu$
$= 4 \times 1.66 \times {10^{ - 27}}$
Now, $E = \dfrac{{{h^2}}}{{2m{\lambda ^2}}}$
So, by substituting the values, we will get
$\Rightarrow E = \dfrac{{{{(6.626 \times {{10}^{ - 34}})}^2}}}{{2 \times 4 \times 1.66 \times {{10}^{ - 27}} \times 12 \times {{10}^{ - 12}}}}$
$= 1.722 \times {10^{ - 11}}eV$

Hence, the kinetic energy of $\alpha$ particle is $= 1.722 \times {10^{ - 11}}eV$ .

Note: In case of electrons going in circles around the nucleus in atoms, de Broglie waves exist as a closed loop such that they can exist only as standing waves and can fit evenly around the loop. So, because of this requirement, the electrons in atoms circle the nucleus in particular configurations or states which are known as stationary orbits.