Q1. An insulated lithium surface is irradiated with light of requency 7 x 10^14 Hz and photoelectrons are emitted until it acquires a positive potential of 0.7v. Calculate the work function of lithium in eV. Take h= 6.6 x 10^-34 and e=1.6 x 10^-19"
I don't get the "until it acquires a positive potential of 07v" part. Can anyone explain?
A1.
Ok ill try to explain the part which you do not understand. It says that the metal surface will continue to emmit photoelectrons until the metal aquires a positive potential of 0.7V. THis means when the metal has 0.7v the electric field set up by the metal is doing sufficient work on the electrons within the metal to stop them from ejecting from the surface of the metal. This means that at this instance, the kinetic energy of the photoelectrons is equal to the work of the electric field.
E
k = W = qV (which you may have to recall from the prelim. course)
E
k = (1.6x10
-19)(0.7)
= 1.12x10
-19 J
Remember the work function is the minimum amount of energy to eject an electron off the surface of a metal, W
f = hf
o, where W
f is the work function and f
o is the threshold frequency.
So using the formulae, hf = W
f - E
k, you can rearrange to form,
W
f = hf - E
k
= 6.6 x 10
-34 x 7 x 10
14 - 1.12x10
-19
= 3.5x10
-19 J
Not sure bout the eV thingo, but wiki knows all
Wikipedia [url said:
http://en.wikipedia.org/wiki/Electronvolt[/url]]
The electronvolt (symbol eV, or, rarely and incorrectly, ev) is a unit of energy. It is the amount of kinetic energy gained by a single unbound electron when it passes through an electrostatic potential difference of one volt, in vacuum. The one-word spelling is the modern recommendation although the use of the earlier electron volt still exists.
One electronvolt is a very small amount of energy:
1 eV = 1.602 176 53 (14)×10−19 J. (Source: CODATA 2002 recommended values)
So if W = 3.5x10
-19 J
and n eV = n * 1.6x10
-19 J
n J = n ev / 1.6x10
-19
so W = 3.5x10
-19 / 1.6x10
-19 eV
= 2.1875 eV
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Q2. "A metal surface of area 1000mm^2 is irradiated with light of requency 6.5 x 10^14 and intensity 0.5W/m^2. If the saturation photoelectric current is 1.0 micromaps, calculate the chance that a photon ejects an electron from the metal surface"
What is the "saturation photoelectric current"?
A2. "Saturation photoelectric current" means that all the photoelectrons emmited from a metal surface is collected by the collector.
Im so unsure if this is right but im gonna give it my best shot, took me alot of though.
A = 1000 mm^2
= 1x10
-3 m
2
Let i be the intensity
i = 0.5 W m-2
By inspection of the units it can be seen that intensity = Power / Area
So, i = P/A
Power is the rate of energy transfer, so
P = nE/ t
Where E is the energy on one proton, n is the number of protons and t is time.
Remember that the ampere is the rate of flow of coloumbs, so
I = nq/t, where I is the current (amperes), n is the number of electrons flowing in the circuit, q is the charge on one coloumb and t is the time interval.
so rewriting the above, n = It/q
and as the number of protons ejected will equal the amount of protons flowing through the circuit.
hence P = (It/q) * (nE/t)
= EI/q,
so A = EI/iq
= (6.6x10^-34)(1x10^-6)/ (0.5)(1.6x10^-19)
= 8.25x10^-21 m^2
So that means that the beam of light hits the metal surface covering an area of 8.25x10^-21 m^2.
Rewiriting the original area in terms of m^2, A of metal = 1 x 10^-3 m^2
Dont get what to do with this though now...
