Max Wavelength of Electrons Emitted from Zinc Surface

[nathan17]

Homework Statement

The work function of zinc is 4.31eV. What is the maximum wavelength of light that will cause electrons to be emitted from a zinc surface?

Homework Equations

I'm not sure because, we are given the Work Function (W) which is 4.31eV and we need to find the max. wavelength (λ) so we will need a couple equations)

v = fλ (v = speed of light)
W = hf₀ (f₀ = threshold frequency) But we already know the Work Function

The Attempt at a Solution

I'm completely stumped. I have looked through the protons section in the book "Physics Key Ideas Part 2" but have yet to find anything like this. I would be extremely happy if someone could point me in the right direction on what equations I would need to use.

If you would like to see the equation sheet we are given, it can be downloaded here:
http://www.ssabsa.sa.edu.au/science/phys-formula-sheet.pdf

Cheers for any help,
Nathan

 

[alphysicist]

Hi nathan17,

Homework Statement

The work function of zinc is 4.31eV. What is the maximum wavelength of light that will cause electrons to be emitted from a zinc surface?

Homework Equations

I'm not sure because, we are given the Work Function (W) which is 4.31eV and we need to find the max. wavelength (λ) so we will need a couple equations)

v = fλ (v = speed of light)
W = hf₀ (f₀ = threshold frequency) But we already know the Work Function

The Attempt at a Solution

I'm completely stumped. I have looked through the protons section in the book "Physics Key Ideas Part 2" but have yet to find anything like this. I would be extremely happy if someone could point me in the right direction on what equations I would need to use.

If you would like to see the equation sheet we are given, it can be downloaded here:
http://www.ssabsa.sa.edu.au/science/phys-formula-sheet.pdf

Cheers for any help,
Nathan

Did you use the equation you listed in your post? What did you get?

 

[nathan17]

Ok, I think I got it :D

W = hf₀
W/h = hf₀/h
Cancel out the h on the right side
f₀ = W/h
where W = 4.31eV = 4.31 * 1.6x10^-19 = 6.9x10^-19J
f₀ = 6.9x10^-19 / 6.63x10^-34
= 1.04x10¹⁵Hz

So,

v = fλ
v/f = fλ / f
Cancel out the f on the right side
λ = v/f
λ = 3.0x10⁸ / 1.04x10¹⁵
= 2.88x10^-7m

Yay, I got it :D

Checked the back of the book and that's the answer, assuming the book is correct!

[Edit] Also, I may need some help on another question in a minute, is it ok to post here instead of making a new thread?

 

[alphysicist]

Your procedure looks right to me!

(About another question: to make it more likely to get help, I think it's best to start a new thread if it is a separate problem.)

 

[nathan17]

Cool, thanks!

Yeah, it's a separate question, I will have a crack at it and see how I go and start a new thread if needed.

Cheers

 

[will.c]

A good technique to practice is to use algebraic manipulation to define the variable you're solving for in terms of the variables given in the problem: for example, we have

c=fλ
W=hf

So, f=c/λ, which we substitute into the second equation and get

W=(hc)/λ

Multiply both sides by λ/W and you have your answer

λ = hc/W

hc is a pretty common combination of constants, so it's good to remember that it's about 1240 eV*nm

then, λ is about 1240 eVnm / 4.31 eV, which is about 288 nm; exactly what you found.

 

[nathan17]

A good technique to practice is to use algebraic manipulation to define the variable you're solving for in terms of the variables given in the problem: for example, we have

c=fλ
W=hf

So, f=c/λ, which we substitute into the second equation and get

W=(hc)/λ

Multiply both sides by λ/W and you have your answer

λ = hc/W

hc is a pretty common combination of constants, so it's good to remember that it's about 1240 eV*nm

then, λ is about 1240 eVnm / 4.31 eV, which is about 288 nm; exactly what you found.

Ah, Cool. Thankyou!
I only recently got good at rearranging equations (I have a physics tutor) so I didn't really think of that, makes it a lot easier.

Cheers

 

[Redbelly98]

I only recently got good at rearranging equations (I have a physics tutor)
Cheers

It's a good skill to keep practicing and maintain at every opportunity. I wish more students would use it

Regards,

Mark