Question

. CP (a) A particle with mass mhas kinetic energy equal to three times its rest energy. What is the de Broglie wavelength of this particle? (Hint:You must use the relativistic expressions for momentum and kinetic energy (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in metres) if the particle in part (a) is (i) an electron and (ii) a proton.

Short answer

We get the de Broglie wavelength of this particle:

Therefore the the numerical value of the kinetic energy as an an electron and (ii) a proton.are for wavelength and

Step-by-step solution

About De Broglie wavelength

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.

Determine the de broglie wavelength

From equation the de Broglie wavelength A of a particle with momentum p is given by;

From equation 37.38, the total energy E of a particle is the sum of the kinetic energy and the rest energy of this particle:

From equation 57.39, the total energy of a particle can be expressed in terms of the momentum of the particle and its rest

mass as:

Calculations

(a) We know that the rest energy of a particle is mag; 30, for a particle that has kinetic energy equal to three times its rest

energy, we have:

Substituting into equation (2), we get the total energy of the particle:

Now, we substitute E into equation (3) and solve for p, so we get the momentum of the particle:

_]

_\

Therefore We get the de Broglie wavelength of this particle:_\

Step 3:Determine the numerical value of the kinetic energy as an an electron and (ii) a proton.

(b) (i) For an electron (m, 2 9.111 10““ kg):

\Ve plug our vnhle for m, into equation (4), so we get the kinetic energy of

the electron:

'I‘hen, we plug the value m,_. into 9(‘llmlion (5), so we gel. the wavelength of

4.119. p]m‘-.i.rnn- ?ii?

'l‘hen, we plug the value m,_. into «111011011 (5), so we get. the wavelength of

the 9100mm:

(ii) For a proton (

\Ve plug our value for m 1, into equation (4), so we get the kinetic energy of

the proton:

_\

Therefore the the numerical value of the kinetic energy as an an electron and (ii) a proton.are_\\for wavelength and_\