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PLANCKS CONSTANT IN THE LIGHT OF AN INCANDESCENT LAMP
SOLUTION
TASK 1
Draw the electric connections in the boxes and between boxes below.

a)
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t 24 sC |
T 297 K |
DT = 1 K |
b)
V /mV |
I / mA |
RB /W |
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Vmin 9.2 mV * This is a characteristic of your apparatus. You cant go below it. |
We represent RB in the vertical axis against I.

In order to work out RB0 , we choose the first ten readings.
c)
V /mV |
I / mA |
RB /W |
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|

Error for RB (We work out the error for first value, as example).
We have worked out RB0 by the least squares.

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RB0 11,4 W |
D RB0 W |
d) ![]()
Working out the error for two methods:
![]()
Method B
Higher value of a:
![]()
![]()
Smaller value of a: ![]()
![]()
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a = |
Da = |
Because of 2Dl = 620 565 ; Dl = 28 nm
|
l = 590 nm |
Dl = 28 nm |
a)
V /V |
I / mA |
R /kW |
b)
Because of ![]()
For working out Dg we know that:
R DR 0.01 kW
R DR 0.01 kW
Transmittance, t = 51.2 %
Working out the error for two methods:
![]()
Higher value of g ![]()
Smaller value of g ![]()
![]()
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R = 5.07 kW |
g = 0.702 |
|
R = 8.11 kW |
Dg = 0.005 |
c)

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d)
V /V |
I / mA |
RB W |
T / K |
RB-0.83 (S.I.) |
R / kW |
ln R |
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unnecessary | ||||||
We work out the errors for all the first row, as example.
Error for RB:
Error for T:
Error for RB-0.83 :

Error for lnR :
e)
We plot ln R versus RB-0.83 .


Because of
![]()
and
![]()
![]()
then
![]()

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h = 6.3 10-34 J s |
D h 10-34 J s |
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