How to find the binding energy of a photon?

AI Thread Summary
To find the binding energy of an electron when an X-ray photon strikes a surface, the initial calculations involved converting the kinetic energy of the emitted electron from eV to joules. The photon energy was also calculated using its wavelength. The confusion arose when attempting to convert the binding energy to kJ/mol, where the user initially overlooked the need to incorporate Avogadro's number. After recognizing this, the correct approach involved converting the energies properly before calculating the binding energy for one mole of electrons. Ultimately, the user successfully solved the problem by making the necessary adjustments.
Mayhem12
Messages
3
Reaction score
0
An X-ray photon of wavelength 0.940 nm strikes a surface. The emitted electron has a kinetic energy of 947 eV. What is the binding energy of the electron in kJ/mol?
What I did:

(947 eV)(1.602E-19 J)/1 eV = 1.52E-16 J(6.626E-34 J.s)(3.0E8 m/s)/.940 nm (10^-9m/1 nm) = 2.11E-16 J

2.11E-16 J - 1.52E-14 J/1000 KJ/J/1 mol = 5.90E-20 Kj/molObviously, there is something terribly wrong here and my textbook is deemed useless at this point. PLEASE HELP! Thanks
 
Physics news on Phys.org
Welcome to PF.
Mayhem12 said:
An X-ray photon of wavelength 0.940 nm strikes a surface. The emitted electron has a kinetic energy of 947 eV. What is the binding energy of the electron in kJ/mol?



What I did:

(947 eV)(1.602E-19 J)/1 eV = 1.52E-16 J


(6.626E-34 J.s)(3.0E8 m/s)/.940 nm (10^-9m/1 nm) = 2.11E-16 J

2.11E-16 J - 1.52E-14 J/1000 KJ/J/1 mol = 5.90E-20 Kj/mol


Obviously, there is something terribly wrong here and my textbook is deemed useless at this point. PLEASE HELP! Thanks
You have found the binding energy in kJ for a single electron, not for one mole of electrons.
 
Okay... so what am I missing? Avagadro's number maybe?
 
Mayhem12 said:
Okay... so what am I missing? Avagadro's number maybe?
Yes.
 
I solved it. I had to convert before doing the E-photon and E-released.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top