Classical physics wavelength of monochromatic light

[Stevee]

Homework Statement

A stopping potential of 0.50 V is required when a phototube is illuminated with monochromatic light of wavelength 600 nm. Monochromatic light of a different wavelength is now shone on the tube, and the stopping potential is measured to be 1.1V. What is the wavelength of this new light? (c = 3.00 × 108 m/s, e = - 1.60 × 10-19 C, h = 6.626 × 10-34 J · s, 1 eV = 1.60 × 10-19 J)

Homework Equations

E=h*f (f=c/lambda)
h*f=sigma+Ek

The Attempt at a Solution

E=h*c/lambda
lambda=h*c/E

I am just unsure on how to obtain the energy for the new monochromatic light given another.
Thankyou!

 

[andrevdh]

How do you incorporate the stopping potential in the equation?
What does it tell you about the problem?

 

[Stevee]

So, the stopping potential would be equivalent to the available kinetic energy which is [Avail. Ek=E - sigma] right? Do i then work out the work function from the other provided wavelength?
The stopping potential tells us the amount of voltage required to stop the movement of electrons, so into the tube?

 

[andrevdh]

The stopping potential is used to bring the emitted photoelectrons to a halt.
How can you calculate the kinetic energy of the photoelectrons from this?

 

[Stevee]

Oh, sorry photoelectron not electron.. my mistake
well I am not entirely sure, I am having a real blank at the moment... I am not sure on whether its possible to use the first wavelength (and stopping pot.) to calculate the work function to sub into the equation, but the frequency changes with wavelength and therefore so does the kinetic energy, maybe I am overthinking this?

 

[andrevdh]

What is the relationship between electric charge and voltage?

 

[Stevee]

q=E/V?
oh woww.. haha
You use E=qV to find the energy then just sub it into E=h*c/lambda and solve for lambda right?..

 

[andrevdh]

Yes, the electric potential difference or voltage between two points tells us the amount
of energy expended by or given to each coulomb of charge moving between the two points.

 

[andrevdh]

Sorry, I meant to say that the calculation gives you the kinetic energy of the
emitted photoelectrons, not the energy of the photons, but your equation is correct.

 

[Stevee]

Thankyou,
hmm it doesn't seem to be working out for me.
So i used E=q*V
E=(1.6x10^-19)*(1.10)=1.76x10^-19
into... E=h*c/lambda
1.76x10^-19=(6.63x10^-34)*(3x10^8)/lambda
where lambda comes out as113x10^-8m.
im doing something wrong

 

[andrevdh]

The voltage tells you something about the photoelectrons, not the photons!
Since it is stopping the photoelectrons, what do you think it is a measure of?

 

[Stevee]

so it would be equivalent to the energy required to emit a photoelectron from the tube?

 

[andrevdh]

No, the energy you calculated is what a photoelectron loses when it
goes through that potential, so it is its kinetic energy as it leaves the material.

 

[Stevee]

Oh ok, ofcourse haha, so from this we can work out the photon energy?

 

[andrevdh]

Set the equations up for the two different photons

hf = ∑ + E_K

 

[Stevee]

600nm..
[6.64x10^-34*3x10^8]/600x10^-9 = ∑ + 1.76x10^-19
unknown nm...
[6.64x10^-34*3x10^8]/ lambda = ∑ + 1.76x10^-19

 

[andrevdh]

It is easier to see what is going on if you do this

hf₁ = ∑ + E_K1

hf₂ = ∑ + E_K2