[VCE] FREE lectures this July. Places booking out fast. Tell your friends and book now!

June 17, 2019, 09:31:30 pm

### AuthorTopic: Dtermine the number of electrons...  (Read 621 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### #1procrastinator

• Guest
##### Dtermine the number of electrons...
« on: September 21, 2012, 10:27:54 am »
0
A 500W lamp directs a beam of yellow light of wavelength 580nm onto a perfect reflecting surface of area 4.0cm^2

If the electrons in the beam each have the same energy as a photon of yellow light
a) determine the momentum of each electron
b) calculate the de Broglie wavelength of each electron
c) determine the number of electrons that are incident on the surface each second in order to produce a pressure of 0.553N m^-2

I've gotten a and b but can't get the right answer with c). To find the force per electron, I divided the momentum in a) by 1 second and used that to find the number of electrons required to product the required force per square metre but my answer is way off

This is question 5 on page 459 of the Heinemann 12 text by the way

#### Aurelian

• Victorian
• Posts: 585
• Respect: +79
• School: Melbourne Grammar School
##### Re: Dtermine the number of electrons...
« Reply #1 on: September 21, 2012, 01:48:18 pm »
0
Sorry... bit confused by the question; the beam of electrons? You only refer to a beam of yellow light in the question, which is made up of photons...
VCE 2010-2011:
English | Philosophy | Latin | Chemistry | Physics | Methods | UMEP Philosophy
ATAR: 99.95

2012-2014: BSc (Chemistry/Philosophy) @ UniMelb

Currently taking students for summer chemistry and physics tutoring! PM for details.

• Honorary Moderator
• ATAR Notes Legend
• Posts: 3174
• Respect: +320
##### Re: Dtermine the number of electrons...
« Reply #2 on: September 21, 2012, 04:01:22 pm »
+3
I had a look at the text, it continues on from Question 4 which has the key piece of information about the electron beam that's directed onto the 'reflecting surface'.

So the electrons in the beam have the same energy as a photon of yellow light --> $E = \frac{hc}{\lambda} = \frac{hc}{580 \times 10^{-9}} = 3.43 \times 10{-19}J$

a) need to figure out the momentum first before we can do c)

$p = mv$
$E_k = \frac{1}{2}mv^2 \implies v^2 = \frac{2E_k}{m}$
$p^2 = m^2v^2$
$p^2 = m^2\frac{2E_k}{m} = 2E_km$

$p = \sqrt{2E_km} = \sqrt{2 \times 3.43 \times 10^{-19} \times 9.1 \times 10^{-31}} = 7.9 \times 10^{-25} kg m/s$

c) determine the number of electrons that are incident on the surface each second in order to produce a pressure of 0.553N m^-2

So Pressure is the force per sq. metre. $P = 0.553 N m^{-2} = \frac{F}{A}$

$100^2 cm^2 = 1 m^2$

The area is $A = 4 cm^2 = 0.0004 m^2$

So the force acting on all the electrons to achieve that pressure is: $F = PA = 0.553 \times 0.0004 = 2.21 \times 10^{-4} N$

We also know Newton's second law: $F = \frac{\Delta p}{\Delta t}$

If we consider the force that would act for 1 second, and the momentum that would act for 1 second, and assume that each electron feels the same amount of force, pressure etc, then we can say:

$F = k \frac{\Delta p}{\Delta t} = k \times \Delta p$  (where k is just a number representing the number of electrons)

$F = k \times 7.9 \times 10^{-25} kg m/s = 2.21 \times 10^{-4} N$

$k = \frac{2.21 \times 10^{-4} N}{7.9 \times 10^{-25} kg m/s} = 2.8 \times 10^{20}$ electrons.

I think that's the correct reasoning.

edit: hmm, was that what you did as well? I just took a look at what the book had and they had $1.39 \times 10^{20}$.

edit: fixed a typo where I had an extra 'k' in the equation when it shouldn't have been there.
« Last Edit: September 21, 2012, 07:28:49 pm by laseredd »

#### #1procrastinator

• Guest
##### Re: Dtermine the number of electrons...
« Reply #3 on: September 21, 2012, 06:37:54 pm »
0
Woops, sorry for the ambiguity, thanks laseredd for clearing it up.

I got 2.7*10^22 but Im pretty sure i messed up the conversion of cm^2 to m^2...I'm finding that pretty confusing. I found the number of electrons first for a sqm by dividing 0.533N by the force per electron (same answer as yours) and then finding the number for the area required which I thought was 0.04m^2

• Honorary Moderator
• ATAR Notes Legend
• Posts: 3174
• Respect: +320
##### Re: Dtermine the number of electrons...
« Reply #4 on: September 21, 2012, 07:22:00 pm »
+1
Hmm, so the unit conversion is where you're having trouble. I'll have a go at explaining it.

1 m^2 is defined as the area of a square with side lengths of exactly 1 metre.

$A = L^2 = (1m)^2 = 1 m^2$

A square centimetre is a square with side lengths of exactly 1cm.

$A = L^2 = (1cm)^2 = 1 cm^2$

We know that a side length of 1 metre is 100cm, so:
$A = L^2 = (100cm^2) = 10000 cm^2$

We can see that 1 cm^2 isn't the same as 0.01 m^2:

$A = L^2 = (0.01 m)^2 = 0.0001 m^2$

We can also see that 1 m^2 isn't the same as 100 cm^2

$A = (100cm)^2 = 10000cm^2$

So the approach you took was:

Force on one electron is $F_1 = 7.9 \times 10^{-25} N$, the pressure is $P = 0.553 N m^{-2}$

Number of electrons in a square metre: $\frac{P}{F_1} = 7 \times 10^{23}$

Then times that by the area wanted: $7 \times 10^{23} * 0.0004m^2 = 2.8 \times 10^{20}$

#### #1procrastinator

• Guest
##### Re: Dtermine the number of electrons...
« Reply #5 on: September 22, 2012, 04:49:48 pm »
0
Thanks a lot man!