Jacaranda questions (1 Viewer)

Wohzazz

Member
Joined
Nov 24, 2003
Messages
512
Location
Sydney
Gender
Male
HSC
2004
1. A beam of UV light of frequency 7x10^15Hz is indirected towards a metal surface in a circuit. (there's suppose to be a diargram). If the maximum kinetic energy of emitted electrons is 9x10^-19J, calculate the
a) the potential required to stop electrons reaching the collector.
b)the work function of the material on which the light is shining.
c) the threshold frequency of the material.
2. A photon collides wwith anelectron and is scattered backwards so that it travels back along its original path,. Describe and expalin the expected wavelength of the scattered light.
3. One electron is ejected from a clean zinc plate by ultraviolet light has kinetic energy of 4X10^-19
a) What would be the kinetic enrgy of this electron when it reached the anode, if a retarding voltage of 0.9V was applied between anode and cathode?
b) what is the minimum retarding voltage that would prevent this electron reaching the anode?
c) all electrons ejected from the zinc plate are prevrnted from reach9ing the anode by a retarding force of 4.3V. What is the maximum kinetic energy of the electrons ejected from the zinc.
 
Last edited:

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
1.

f = 7*10^15 Hz
K = 9 * 10^-19 J
e = 1.6 * 10^-19 C (electron charge)
h = 6.6 * 10^-34 Js (Planck's const)
m = 9.1*10^-31 kg (electron mass)

a) The stopping voltage V = K/e = 5.63 V

b) The energy contained by each photon is E = hf = 4.62*10^-18 J. The difference in the kinetic energy of the electrons, and the energy of the photons is the work function @. (i.e. E = K + @).

@ = E - K = 3.72*10^-18 J

c) The threshold frequency F is the minumum frequency required to remove electrons from the metal.

@ = hF
F = @/h = 5.64 * 10^15 J

2. This question is to do with the Compton effect where photons collide with electrons and the wavelength of the photon which is scattered, is changed.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/comptint.html

let L1 be the wavelength of the incident photon, let L2 be the wavelength of the scattered photon, let m be the electron mass, @ = angle of scattering, c is the speed of light.

L2 - L1 = h/mc (1 - cos@)

here @ = 180 degrees since it's scattered backwards.

L2 - L1 = 2*h/mc

therefore the wavelength of the scattered photon will be slightly higher than the incident photon, by 2h/mc metres.

3

a)

Applying a retarding voltage V causes the kinetic energy of the electrons to drop by Ve.

So kinetic energy of the electrons,
K = K0 - Ve
= 4*10^-19 - 1.44*10^-19
= 2.56*10^-19 J

b)

In order to stop the electron reaching the anode you need to apply a stopping voltage V such that K = 0.

K = K0 - Ve
V = (K0 - K)/e
= (K0 - 0)/e
= K0/e
= 2.5V

c)

If 4.3V are required to stop the electrons, then what kinetic energy will they have when V = 0?

V = 4.3V stopping voltage causes a kinetic energy drop of Ve = 6.88*10^-19 J

So the maximum kinetic energy of the electrons being ejected from the zinc is:

K = 6.88*10^-19 J
 
Last edited:

Wohzazz

Member
Joined
Nov 24, 2003
Messages
512
Location
Sydney
Gender
Male
HSC
2004
Thanks so much wogboy:) You're so great and explain so well.
But there is one i thing i need to ask, what iss this formula

K=K0-Ve
and what does each symbol stand for. Thanks
 
Last edited:

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
K=K0-Ve
and what does each symbol stand for. Thanks
K = kinetic energy of the electrons (J)
K0 = kinetic energy of the electrons, whenever no retarding voltage is applied (J)
V = retarding voltage (V)
e = electron charge = 1.6*10^-19 C

All it's really saying is that if I apply a retarding voltage V, then the kinetic energy of the electrons is being reduced by Ve. That's all :)
 

Wohzazz

Member
Joined
Nov 24, 2003
Messages
512
Location
Sydney
Gender
Male
HSC
2004
Originally posted by wogboy

All it's really saying is that if I apply a retarding voltage V, then the kinetic energy of the electrons is being reduced by Ve. That's all :)
Just that i haven't seen the equation written down before. It was probably expected that you worked it out yourself. for K0 is that a zero or O? Is that the proper symbol?
 

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
I derived that formula K = K0 - Ve, I'm not sure if you'll find it in books, but I'll show you how I derived it, below.

--------------------------------------

Imagine the anode and cathode of the phototube are x metres apart, there is electromagnetic radiation frequency f coming in, the electron has mass m and charge e, the anode has work function @.

The electron receives kinetic energy (K = hf-@) from the photons. Since you know that K = 1/2 m*u^2 (I'm using u here instead of v),

1/2*m*u^2 = hf - @
u = sqrt [2*(hf - @)/m]
(this is the velocity at which the electrons are "kicked off" by the photons)

Now when you apply a retarding voltage V across the tube, you are setting up an electric field E=V/x, agreed? What happens whenever you have a charge q, in an electric field? A force F = qE acts on it.

F = -eE
= -eV/x

but F = ma (Newton's 2nd law)
ma = -eE
a = -eE/m
= -eV/(mx)

so what's happening is that the electron is being pulled back (or retarded) by the voltage.

to determine the final speed of the electron at which it reaches the cathode.

v^2 = u^2 + 2as
= 2*(hf - @)/m - 2*x*eV/(mx)
= 2*(hf - @)/m - 2*eV/m

so now that we know the final speed at which it hits the anode, let's work out the kinetic energy at this point.

K = 1/2*m*v^2
= 1/2*m*{2*(hf-@)/m - 2*eV/m}
= hf - @ - eV

K = (hf - @) - eV

Whenever you apply no retarding voltage, you know that the kinetic enegy = (hf - @). Call this K0 (k-zero). This is the kinetic energy of the electrons, if you don't apply a retarding voltage.

K = K0 - Ve

-------------------------------------------

Don't worry too much about that long boring derivation, I just want to make the point that if you apply a retarding voltage V, you will lose kinetic energy Ve. It should be obvious intuitively.

The question 3a) you gave me was quite unusual (for a HSC level question) since it's not a simple "memorise formula -> sub in values -> solve" sort of question. You have to sort of formulate your own methods of solving the problem (it's unlikely BOS would put this sort in the exam since all calculation questions I've seen in HSC papers are straightforward and don't give you the opportunity to think broadly). Don't be put off if you don't understand it.
 
Last edited:

Wohzazz

Member
Joined
Nov 24, 2003
Messages
512
Location
Sydney
Gender
Male
HSC
2004
I think i understand how you derived it but i doubt i will remember it. But as you implied, it's not required of me.
I don't think it's in the syllabus but Jacaranda included it. I wonder why considering it's usually the text that adheres most to the syllabus.

Wogbog, you are great at teaching. Are you like tuitoring science or something? Doing any science degrees?
 

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
Currently studying electrical engineering at uni. Not really tutoring, just helping out in my spare time. Thanx for the compliment :)
 

Dash

ReSpEcTeD
Joined
Jul 17, 2003
Messages
1,672
Location
nExT dOoR fOoL!
Gender
Male
HSC
2003
lol yeh I agree Wohzazz.
I used to look forward to wogboy's posts when I had a question back in yr 12 :D
great job woggy... lol i can call u that right? :p
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top