Sapling Question 22

[Keon Amirazodi 3H]

The question is:

You use an electron microscope in which the matter wave associated with the electron beam has a wavelength of 0.0339 nm. What is the kinetic energy of an electron in the beam, expressed in electron volts?

How do we know to use the De Broglie equation to solve for the energy in this question instead of using c=wv and E=hv? And why does it result in different answers depending on which method you use to solve for energy?

When using the De Broglie equation, the energy is 1310 eV while when using the other equation set, the energy would be 36,602 eV.

[Giselle Granda 3F]

You use the de Broglie equation because the electron has mass (and momentum). Therefore you can use the equation lamda=h/mv, to solve for the velocity of the electron since you are given its wavelength and know its mass (9.11x10^-31kg), and with the velocity you can you the equation for kinetic energy. The correct answer would be your first one!

[Lucy Wang 2J]

The reason you must use de Broglie's equation for this problem is because it is talking about an electron. The equation c=wv only applies to photons since c is the speed of light. When you use that equation for an electron, your answer will be incorrect.

However, de Broglie's equation can be used for any moving particle that has mass and since an electron has mass, you would use de Broglie's for this problem.

[HannahRobinson3L]

Hi! The equations of c=wv and E=hv are associated with atomic spectra, which involves an electron either absorbing or emitting energy to either jump or drop to a higher or lower energy level, and using the resulting spectrum to identify elements. This is a completely different question from the one asked in the question. For this question, they are asking what the kinetic energy of an electron in the electron beam is. In this case, the problem is just asking what the kinetic energy of an electron in this electron beam with wavelike properties is, so we know to use De Broglie Equation.