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PLANCK’S CONSTANT IN THE LIGHT OF AN INCANDESCENT LAMP SOLUTION

electronics

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TERMENI importanti pentru acest document

PLANCK’S CONSTANT IN THE LIGHT OF AN INCANDESCENT LAMP

SOLUTION




TASK 1

Draw the electric connections in the boxes and between boxes below.


TASK 2

a)

t 24 sC

T 297 K

DT = 1 K

b)

V /mV

I / mA

RB /W

Vmin 9.2 mV *

This is a characteristic of your apparatus. You can´t go below it.

We represent RB in the vertical axis against I.


In order to work out RB0 , we choose the first ten readings.

TASK 2

c)

V /mV

I / mA

RB /W


Error for RB (We work out the error for first value, as example).

We have worked out RB0 by the least squares.

RB0  11,4 W

D RB0 W

d)

Working out the error for two methods:

Method A

Method B

Higher value of a:

Smaller value of a: 

a =

Da =

TASK 3

Because of 2Dl = 620 – 565 ; Dl = 28 nm

l = 590 nm

Dl = 28 nm

TASK 4

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:

Method A

Method B

Higher value of g

Smaller value of g

R = 5.07 kW

g = 0.702

R’ = 8.11 kW

Dg = 0.005

c)

d)

V /V

I / mA

RB W

T / K

RB-0.83 (S.I.)

R / kW

ln R



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

h = 6.3 10-34 J · s

D h 10-34 J · s






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