Question

What will be the work function of a substance for a threshold frequency of 43.9*10^13 Hz.

Answer 1

@Algebra2013 @genius12 @xxAshxx

Answer 2

Hi! This is photoelectric effect, right?

Answer 3

yes

Answer 4

Well, the threshold frequency is the frequency that has enough energy to get an electron away from the rest of the atom. The highest energy electron will be the easiest to dislodge, and so it will require the least energy to leave. The work function is that energy \(\rightarrow\) the minimum energy needed to eject an electron.

The energy of that frequency, then, is equal to the work function. The energy of a photon is \(h\ f\), and our frequency is threshold frequency - symbolized as \(f_0\).

So, all in all, \(\phi =h\ f_0\).

Answer 5

That is
1. what I found here:http: //physics.info/photoelectric/
and
2. what I put together from the definition of the work function - and then put into math.

Answer 6

Any questions?

Answer 7

@theEric solve this by putting values

Answer 8

Oh yeah, \(h\) is Planck's Constant. There is a table of its values in different units here.

I'm used to using \(6.626\times 10^{-34}\ [J\ s]\) If you multiply it with the frequency (which is in \([Hz]=\dfrac{1}{[s]}\)), your work function will be in joules.

Answer 9

Ah, I never included the link for the values of \(h\).
http://en.wikipedia.org/wiki/Planck_constant

Answer 10

I won't solve it for you, that's up to you!
\(\phi=h\ f_0\)
\(f_0=43.9\times 10^{13}\ [Hz]\)
\(h\approx 6.626\ [J\ s]\)

Answer 11

@theEric the anwer is 1.82eV

Answer 12

I agree.

Answer 13

:)

Answer 14

My Windows calculator got me \(1.815704\ [eV]\), and your answer has proper significant figures.

Answer 15

@theEric thanks for solving

Answer 16

Haha, you did the solving :)