# Linera Expansion and heating from a stream of electrons

1. Mar 21, 2009

### TFM

1. The problem statement, all variables and given/known data

A thin square tungsten foil, 10mm x 10mm, is supported in the centre of a vacuum chamber by thin wires which have negligible thermal conductance. A cooled shield surrounds the foil such that its initial temperature is -37 C. Two parallel slits, 1$$\mu$$m wide and 1mm long and whose centres are separated by 6.6$$\mu$$m, have been cut in the foil. The foil is now heated by electron bombardment with a 10mA current of 5keV electrons, whose energy is completely absorbed by the foil. Calculate the change in separation of the slits.

How would you go about measuring this change in separation from outside the vacuum chamber?

For tungsten, the emissivity should be taken as (1/5.7) and the coefficient of linear expansion as 5 x $$10^{-6} K^{-1}$$. (The linear expansion coefficient $$\alpha$$, is defined as the fractional change in linear dimension per degree Kelvin, i.e. $$\alpha = \DeltaL/L0$$ where $$\Delta$$L is the change in length for a 1K change in temperature and L0 is the original length.)

2. Relevant equations

$$\frac{\Delta L}{L_0} = \alpha \Delta T$$

3. The attempt at a solution

Okay, so to get the change in length, Delta L, I need the formula

$$\Delta L = L_0 \alpha\delta T$$

However, I am unsure how to get T.

When I saw that there was a current, I thought of using the power law

P = VI
and

V=IR

but we are not given any resistance. we are given the energy of each electron, But I am unsure how to get through to the current.

TFM

2. Mar 21, 2009

### Redbelly98

Staff Emeritus
Since you have V and I, you can calculate P. R is not needed.

3. Mar 22, 2009

### TFM

Where are we given the voltage? I know the electrons have 5 keV, but I thought this was an energy not a voltage...?

4. Mar 22, 2009

### Redbelly98

Staff Emeritus
True, 5 keV is an energy.

What voltage would give an electron an energy of 5 keV?

5. Mar 22, 2009

### TFM

well, 1 electron volt is the energy to get a electron through a p.d. of 1 volt,

so there is a voltage of 5kV then.

so now,

$$VI = P$$

$$P = 5000 * 10*10^-3 = 50 Watts$$

So this is the energy per second

so do we need to find out how long the bombardment lasted?

6. Mar 22, 2009

### Redbelly98

Staff Emeritus
Yes, 50W is right. Good.

Good question.

They don't say anything about how long it lasts. The do tell us the emissivity of tungsten. That's a hint that radiation is playing a role here.

7. Mar 22, 2009

### TFM

$$j = \sigma T^4$$

j = power over area, therefore,

$$T^4 = \frac{P}{a\sigma}$$

However, the emissivity is the ratio of energy emitted by object compared to the energy emitted by a perfect Blackbody.

therefore

$$1/5.7j = \sigma T^4$$

$$T^4 = \frac{P}{5.7 a\sigma}$$

So this gives:

$$T^4 = \frac{50}{5.7 (0.01*0.01 - 2(0.001*0.000 001))\sigma}$$

$$T^4 = \frac{50}{5.7 (0.0001 - 0.000000002)\sigma}$$

$$T^4 = \frac{50}{5.7 (0.000099998)\sigma}$$

$$T^4 = \frac{50}{3.23*10^{-11}}$$

$$T^4 = 1.55*10^{12}}$$

$$T = 1115K = 841 C$$

Does this look right?

8. Mar 22, 2009

### Redbelly98

Staff Emeritus
Pretty close! Two little errors to correct:

j = ε σ T4 = (1/5.7) σ T4 A real object radiates less than a blackbody

Also, the square foil has 2 sides, so total surface area is ___?

9. Mar 22, 2009

### Redbelly98

Staff Emeritus
p.s. I'm logging off the forum for a while ... good luck!

10. Mar 22, 2009

### TFM

Oka the area of the foil plate is 10mm x 10mm = 0.0001m^2

However, the two cuts have a total area of 0.000000002 m^2

Thus the total area is 0.000099998 m^2

So for both sides,

Area = 0.000199996m^2

$$P = A\epsilon \sigma T^4 [\tex] [tex] \frac{P}{A\epsilon \sigma} = T^4 [\tex] [tex] \frac{50}{0.000199996 * \frac{1}{5.7}*5.67*10^{-8}} = T^4 [\tex] [tex] T^4 = 2.5 * 10^{13}$$

$$T = 2239K = 1966 C$$

So I presume this is the final Temperature, so

$$\Delta L = L_0 \alpha\delta T$$

$$\Delta L = L_0 \alpha (T_f - T_i)$$

$$\Delta L = 6.6*10^{-6} *5*10^{-6} (1966 - 37)$$

Gives a change in length of 6.37 * 10^-8 (657 micro meters)

Does this look okay?

11. Mar 22, 2009

### Redbelly98

Staff Emeritus
Very close!

If you subtract -37 from 1966 °C, that's the same as adding ____ °C.

6 x 10-8 m is how many μm? Or, how many nm?

p.s. It's probably okay to neglect the area of the slits, since many quantities are given to just 2 significant figures.

12. Mar 22, 2009

### TFM

Drat, I forgot it was minus 37. Thankfully, this is where Excelcomes in handy, just change one sign.

Gives 6.6 * 10^-8m = 66 nanometers

13. Mar 22, 2009

### Redbelly98

Staff Emeritus
Looks good

14. Mar 22, 2009

### TFM

Excellent. I now need to figure how to measure the difference. I am assuming a ruler won't be of any use (How would you go about measuring this change in separation from outside the vacuum chamber?)

This seems though such a small change in seperation. Would a laser be useful?

15. Mar 22, 2009

### Redbelly98

Staff Emeritus
Yes, a laser would be extremely useful.

Are you aware of an experiment that involves a laser -- or any single-wavelength source of light -- and two closely-spaced slits?

16. Mar 23, 2009

### TFM

The Youngs Double Slit experiment.

As the slits become wider, the destructive and constructive interference spots will move.

17. Mar 23, 2009

### Redbelly98

Staff Emeritus
Yes, exactly.

18. Mar 23, 2009

### TFM

Excellent.

Well, Thanks a lot for your assistance, Redbelly98