What is the energy of an X-ray photon with a wavelength of 6.6*10^–11 m?

AI Thread Summary
The discussion focuses on calculating the energy of an X-ray photon with a wavelength of 6.6 *10^–11 m, which corresponds to a characteristic spectral line of molybdenum. The relevant equations include E=h*f and E=(h*c)/λ, with h being Planck's constant and c the speed of light. The confusion arises from the use of the formula E(ev)=(h*c)/(λ*e), where e represents the elementary charge, leading to the energy being expressed in electronvolts (eV) instead of joules. This conversion is necessary as physicists commonly use eV as a unit of energy, with 1 eV equating to the energy gained by an electron when accelerated through a potential difference of 1 volt. Ultimately, the energy of the X-ray photon is deduced to be 19 keV.
edwinem
Messages
5
Reaction score
0

Homework Statement


The X-ray spectrum of molybdenum has a particular characteristic spectral line of
wavelength 6.6 *10^–11 m. The ionisation energy of molybdenum is 20 keV.

Deduce that the energy of an X-ray photon of wavelength 6.6*10^–11 m is 19 keV.


Homework Equations


E=h*f
E=(h*c)/l
h=planks constant
c=speed of light
l=lambda(wavelength)


The Attempt at a Solution


Normally this is quite easy. You just plug in wavelength into the equation and it should give you the answer. However, the part that I do not understand is that in the markscheme the proper formula to use is E(ev)=(h*c)/(l*e). Were e is elementary charge. Now this does not make sense to me as (h*c)/(l*e)=V (Voltage). So could someone explain why I need to use elementary charge.
 
Physics news on Phys.org
Physicists often use eV as unit for energy instead of joule. 1 eV is the energy an electron gains when it is accelerated across 1 V potential difference. This energy is equal to 1( V ) * e (charge of electron) = 1.6 10-19 J.

ehild
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top