How much energy is carried by one quantum of these electromagnetic waves?

[mustang]

Problem 6.
Radiation emitted from human skin reaches its peak at wavelength=960um.
How much energy is carried by one quantum of these electromagnetic waves? Answer in eV.
I have found that the frequency is 3.125*10^11 Hz.
Note: What do I do?

Problem 8. Light of wavelength 350 nm falls on a potassium surface, and the photoelectrons have a maximum kinetic energy of 1.3 eV.
What is the work function of potassium? Answer in eV.

Problem 17.
Light of wavelength 3*10^-7 m shines on the metal lithium, iron , and mercury which have work functions of 2.3 eV, 3.9 eV, and 4.5eV, respectively.
For those metals that do exhibit the photoelectric effect, what is the maximum energy of the photoelectrons?
where do I start? :happy:

 

[ShawnD]

Problem 6.
Radiation emitted from human skin reaches its peak at wavelength=960um.
How much energy is carried by one quantum of these electromagnetic waves? Answer in eV.
I have found that the frequency is 3.125*10^11 Hz.
Note: What do I do?

Use Plack's Constant.

Problem 8. Light of wavelength 350 nm falls on a potassium surface, and the photoelectrons have a maximum kinetic energy of 1.3 eV.
What is the work function of potassium? Answer in eV.

Calculate the frequency, use Planck's constant to find the energy. Subtract 1.3eV from the energy you calculated, that is the work function.
Work function is the energy required to pull an electron from an atom and have 0 kinetic energy. For these equations, it goes like this:
light energy = work function + photoelectric kinetic energy.

Problem 17.
Light of wavelength 3*10^-7 m shines on the metal lithium, iron , and mercury which have work functions of 2.3 eV, 3.9 eV, and 4.5eV, respectively.
For those metals that do exhibit the photoelectric effect, what is the maximum energy of the photoelectrons?
where do I start? :happy:

Calculate the frequency, calculate the energy with Planck's constant. Take that energy and subtract 2.3 eV for lithium. Subtract 3.9 eV for iron. Subtract 4.5 eV for mercury. Do each of those subtractions separately. If the resulting number is negative, photoelectrons do not come from that metal. Due to the wording of the question ("For those metals that do exhibit the photoelectric effect"), I would suspect mercury does not let go of electrons. That's just suspicion though.

 

[mustang]

In regards for problem 6.

So for problem 6.
I would use the Plack's formula: E=hf
Where the f=3.125*10^11Hz and the h=6.63*10^-34 and solve for E?

 

[swansont]

So for problem 6.
I would use the Plack's formula: E=hf
Where the f=3.125*10^11Hz and the h=6.63*10^-34 and solve for E?

Yes. And then convert the answer, in J, to eV.

 

[mustang]

Problem 6.

For problems 6 I multiplied f with h and got 2.071875*10^-22 .If this is right what is the number (1.6*10^-6, I think) to convert this answer into eV.

 

[swansont]

For problems 6 I multiplied f with h and got 2.071875*10^-22 .If this is right what is the number (1.6*10^-6, I think) to convert this answer into eV.

1 J = 1.6e-19 eV

 

[gnome]

Correction:
1 eV = 1.6 \times 10^{-19}J