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Photoelectric Effect question

Electromagnetic Waves of wavelength 2.5*10^-7 m are incident on the surface of a metal whose work function is 2.8*10^-19 J.
a) Calculate the energy of a single photon.
b) Calculate the maximum kinetic energy of electrons released from the metal
c) Determine the maximum speed of emitted photoelectrons.

It seems everything is dependent on part a being correct, the answers at the back say a) Is 8.3*10^-19 but when I used E=hf and f=C/wavelength I got 1.04975*10^-18 as the answer which is high so I decided to subtract the work function from it but then it became 7.9675*10^-19 which is less than the answer, did the textbook round of as it go, what did I do wrong?

Reply 1

FizzicsGuy

When doing part a i got 7.956*10^-19J but i never subtracted the work function like you said i just used E=hf and v-f(lambda)

:confused:

I think the answer is wrong as far as my knowledge goes. Higher Physics

sorry i don't know how to do equations on tsr.

(edited 11 years ago)

Reply 2

Primus2x

Original post by GreigM

When doing part a i got 7.956*10^-19J but i never subtracted the work function like you said i just used E=hf and v-f(lambda)

:confused:

I think the answer is wrong as far as my knowledge goes. Higher Physics

sorry i don't know how to do equations on tsr.

Substitute f=(3.8*10⁸)/(2.4*10^-7) into E=hf
and I got 1.04975*10^-18 subtract the work function and I get 7.6975*10^-19 so the answer at the back (8.3*10^-19) is wrong?

Reply 3

FizzicsGuy

Original post by Primus2x

Substitute f=(3.8*10⁸)/(2.4*10^-7) into E=hf
and I got 1.04975*10^-18 subtract the work function and I get 7.6975*10^-19 so the answer at the back (8.3*10^-19) is wrong?

What do you use for your value of h?

Reply 4

Asklepios

Original post by Primus2x

Substitute f=(3.8*10⁸)/(2.4*10^-7) into E=hf
and I got 1.04975*10^-18 subtract the work function and I get 7.6975*10^-19 so the answer at the back (8.3*10^-19) is wrong?

Does the wavelength = 2.4 x 10^-7or 2.5 x 10^-7 m? In the OP you have 2.5, but here you have 2.4. Using 2.4 gives the right answer:

{E=\frac{hc}{\lambda}}

{E=\frac{(6.63\times10^{-34})(3\times10^8)}{2.4\times10^{-7}}}

{E=8.29\times10^{-19}}J

In general to find the total energy of a photon, do not subtract the work function. Only subtract the work function when calculating the kinetic energy.

(edited 11 years ago)

Reply 5

AeroPlane04

For a), you can either:
use v=f(lambda) to find f first, and then use E=hf with this calculated value of f.
or substitute f=v/(lambda) into E=hf to get E=hc/(lambda), and then use the value given in the question.
The calculated value of E will be the energy of a single photon.

Reply 6

FizzicsGuy

Original post by Asklepios

Does the wavelength = 2.4 x 10^-7or 2.5 x 10^-7 m? In the OP you have 2.5, but here you have 2.4. Using 2.4 gives the right answer:

E=\frac{hc}{\lambda}

E=\frac{(6.63\times10^{-34})(3\times10^8)}{2.4\times10^{-7}}

E=8.29\times10^{-19}

In general to find the total energy of a photon, do not subtract the work function. Only subtract the work function when calculating the kinetic energy.