Work function of a cathode, when the wavelength of light is increased by 50%

[Tina20]

Homework Statement

The maximum kinetic energy of photoelectrons is 3.14 eV. When the wavelength of the light is increased by 50.0 %, the maximum energy decreases to 1.11 eV. What is the work function of the cathode?

Homework Equations

Kmax = E elec - E not
= hf - hfnot

hf = hc/lambda

*Enot is the work function

Kfinal-Kinital = hf - hfnot

The Attempt at a Solution

Kf = 1.11 eV
Ki = 3.14 eV

change in K = -2.03

From there, I do not know what to do. I don't know how I would go about finding the work function (hfnot)

Can someone please tell me what I should be doing/equating? Your help would be greatly appreciated :)

 

[Tina20]

So i thought of converting the equation I posted below to:

Kmax = hc/lambda - hc/lambda not

because hf = hc/lambda

So if Kmax = -2.03
and h*c = 1.989x10^-25

Then the equation would be as follows:

-2.03 eV = 1.989x10^-25/lambda - 1.989x10^-25/lambda not

now in the question it states that the wavelength was increased by 50%. So I tried to solve the equation with lambda = 1.50 lambda and lambda not staying as lambda.

So the equation then looked like:

-2.03 eV = 1.989x10^-25 (1/1.5 lambda - 1/ lambda)

I converted -2.03 eV * 1.60x10^-19 J/ 1ev = 3.248x10^-19 J ( I decided to take out the negative because my final answer would have negative energy which wouldn't make sense.

3.248x10^-19 J = 1.989x10^-25 (1/1.5 lambda - 1/ lambda)
1632981.398 = 1/1.5 lambda - 1/lambda
1632981.398 = 1/0.5 lambda
1.22475x10^-6 = lambda

Enot = hc/lambda
Enot = 1.624x10^-19 J

This answer was incorrect. Can someone please reply and provide some guidance. Thank you!

Tina

 

[Delphi51]

Good morning, Tina.
I'm used to W for the work function: E = hc/λ - W.
Write this once for the first wavelength: E1 = hc/λ - W
and again for λ increased to 1.5λ: E2 = hc/(1.5λ) - W
You have two equations with two unknowns, λ and W.
Solve one equation for λ and sub into the other to find W.
I get about 3 times the answer you have.

 

[Tina20]

Ok, thank you for your help, I will work on it right now :)

 

[Tina20]

So I solved for lambda and got an answer of 3,265962.795 m for lambda

I substituted that into the equation:

E2 = hc/1.5 lambda - W

I got an answer of 1.776x10^-19 J for -W.

I inputted with positive value of the answer into the computer and it says the answer is wrong. Is my answer above what you roughly got?

 

[Tina20]

Nevermind! I made a mathematical error. I solved the question and got the right answer! Thank you so much!

I posted another post in regards to a proton being fired towards on Oxygen nucleus and I need to find the velocity at which the proton needs to be fired towards the oxygen nucleus. I think I equated my equations wrong, but if you could take a look at my post and give me any pointers I would really appreciate it :)

Thank you Delphi51