What is the Photoelectric Effect?

[pugtm]

Homework Statement

When light of wavelength 298 nm is incident on potassium, the emitted electrons have
maximum kinetic energy of 1.65 eV.

A)What is the energy of an incident photon?
The value of h c is 1240 eV · nm .
Answer in units of eV.

B)What is the work function for potassium?
Answer in units of eV.

C)
What would be the maximum kinetic energy
of the electrons if the incident light had a
wavelength of 375 nm?
Answer in units of eV.

D)
What is the threshold wavelength for the photoelectric effect with potassium?
Answer in units of nm.

Homework Equations

E=H*V
E=H*F

The Attempt at a Solution

A)
this is the part that is confusing, it gives us the HC which should mean that i have the energy of the incident photon according to law of conservation of energy. but surely it can't be that simple, this is an extra chapter we didn't get to so i have to go by what i find online. according to http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html it is a totally different number
any assistance would be appreciated

 

[rock.freak667]

if E=hf and f=c/λ then E = hc/λ .

They gave you hc = 1240 eVnm

and you have λ = 298 nm, then if you put that into the formula, you will get E, the energy of an incident photon.

 

[da_nang]

The value for hc, Planck's constant times speed of light in vacuum, is 4.135669*10^-15 eVs * 2.99792458*10^8 m/s = 1.23984...*10^-6 eVm ~ 1240 eV*nm. So, there's nothing wrong there.

As to part a, it is quite simple if you remember that E = hf for a photon, where f is the frequency of the EM wave/photon.

 

[kamas]

Hi pugtm,
Don't freak out if problems occasionally seem easy - sometimes they are!

Knowing that h*c=1240 eV*nm,
(1240 eV*nm)/(298 nm) = 4.2 eV


Let's check this answer using a different method:

So you know that the wavelength of an incident photon is 298 nm, and we can determine the energy of that photon using the equation Energy = h*nu, where nu is the frequency in s^-1. To convert nm to s^-1, use the speed of light:

(3E10 cm/s)/(298E-7 cm) = 1.01E15 s^-1

E = (6.626E-34 J*s)*(1.01E15 s^-1) = 6.67E-19 J

1 eV = 1.6E-19 J

(6.67E-19J)*(1 eV / 1.6E-19 J) = 4.2 eV

Same thing! I hope this helps.

Cheers,
Kamas

 

[pugtm]

thanks all, what are the equations for the work function and threshold wavelength though?

 

[rock.freak667]

thanks all, what are the equations for the work function and threshold wavelength though?

the energy of the incident photon goes into releasing the electron (work function) + maximum kinetic energy of the electron.