## Problem 1.33

Anna Lapuos 3C
Posts: 10
Joined: Fri Sep 29, 2017 7:07 am
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### Problem 1.33

Why is work function = 16.565 x 10^-18 J? That is, is the work function always going to be attained by multiplying Planck's constant (h) with frequency ?

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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### Re: Problem 1.33

Hi Anna,

The work function is a property of the metal from which an electron is emitted so different metals can have different work functions, or thresholds that an incoming photon must overcome. Energy of the incoming photon is calculated by frequency multiplied by Planck's constant. If the energy of the photon is greater than the work function, an electron will be emitted with some kinetic energy, following the conservation of energy.

$E_{photon} - \phi = E_{kinetic}$

Where $E_{photon} = hv$
and $E_{kinetic} = \frac{1}{2}mv^{2}$

Hope this helps!

Jonathan Tangonan 1E
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

### Re: Problem 1.33

To add on, I think something important to note is the line

" No electrons are emitted from the surface of the metal until the frequency of

This information gives us what we would call the threshold frequency or the minimum frequency required to eject electrons. Just as the work function is the required amount of energy to emit electrons we can say that the value of the Energy of the Photon "hv" and the work function will be the same amount. It is only because we are given the minimum frequency required to eject electrons that we can multiply Planck's constant by the minimum frequency that we get the value of the work function in this case. If we were given a different frequency, we would also need to know another amount like the Kinetic Energy to determine the value of the work function.

Ek=hv - work function
Ek=0 (This is because the Energy of the Photon and the work function are the same in this case)
0 = hv - work function
work function = hv