Question

calucate the de-broglie wavelength of an Alpha particle of kinetic energy 10kev.

Answer 1

HELLO THERE!

We know, that 1 keV = 1000 eV

So, here, Kinetic energy = 10000 eV

Also, 1 eV = 1.6 x 10⁻¹⁹ J

So, given Kinetic energy of the particle = 1.6 x 10⁻¹⁹ x 10⁴ J = 1.6 x 10⁻¹⁵ J

Again,

$KE = \frac{1}{2}mv^{2}$

Where m is the mass of the particle and v is the velocity with which the particle is moving.

An alpha particle is a doubly charged Helium atom, which means that it has two protons and two neutrons.

Mass of one proton = (approximately) Mass of one neutron.

So, mass of an alpha particle = 4 x 1.67 x 10⁻²⁷ kg

= 6.68 x 10⁻²⁷ kg.

Now, put the value of mass m and Kinetic energy KE in the equation to get the velocity with which it is travelling.

$KE = \frac{1}{2}mv^{2}\\\\\implies 1.6 \times 10^{-15} = 6.68 \times 10^{-27} \times v^{2}\\\\\implies v^{2} = \frac{1.6\times10^{-15}}{6.68\times10^{-27}}\\\\\implies v^{2} = 2.39 \times10^{11}\\\\\implies v = 4.88\times10^{5} ms^{-1}$

So, we now have v, m, and h.

h = Plank's constant = 6.626 x 10⁻³⁴ Js

Put these values in the de-Broglie equation:

$\lambda = \frac{h}{mv}\\\\\implies \lambda = \frac{6.626\times10^{-34}}{6.68\times10^{-27}\times4.88\times10^{5}}\\\\\implies \lambda = 2.032\times10^{-13} m$

THIS WAS YOUR ANSWER..

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