1) Using data from two experiments where sodium was illuminated with different wavelengths of light, the work function of sodium was calculated to be 2.28 eV. 2) The stopping potentials measured in the experiments were used along with the speed of light and electron charge to determine the work function and Planck's constant. 3) By setting the equations for the two experiments equal to each other and solving, the work function of sodium was found to be 2.28 eV and Planck's constant was determined to be 6.626×10-34 J·s.

1. 1
When sodium metal is
illuminated with light of
wavelength 4.20 × 102
nm, the
stopping potential is found to
be 0.65 V; when the
wavelength is changed to 3.10
× 102
nm, the stopping
potential is 1.69 V. Using only
these data and the values of
the speed of light and the
electronic charge, find the work
function of sodium and a value
of Planck’s constant.
Helping Tools
No.1

2. 2
No.2
No.3
Solution
max
We know that maximum KE of
photoelectrons is given by
where is the stopping potential.
(1)
s
s
s
hc
K eV hf
V
hc
eV
 



= = − = −
 = − − − − − −

3. 3
2 9
2 9
7 7
In accordance with given data, eq.(1)
takes the forms as:
0.65 (2)
4.20 10 10
1.69 (3)
3.10 10 10
subtracting eq.(2) from eq.(3), we get
1.69 0.65
3.10 10 4.20 10
hc
eV
X X m
hc
eV
X X m
eV eV
hc hc
X m X m


−
−
− −
= − − − −
= − − − −
−
= −
2
4 1
1 1
1.04 [ ]
310 420
420 310
1.04 [ ]
310 420
110
1.04 [ ]
130200( )
1.04eV 8.45 10 ( )
eV hc
nm nm
nm nm
eV hc
nmX nm
nm
eV hc
nm
hcX X X nm
− −
 = −
−
 =
 =
 =
4
9 4
8
13
13
13
15
1.04 1
10
8.45
1.04 1
10 10
8.45
2.998 10
1.04 . 1
10
2.998 8.45
1.04 . 10
0.041 10 .
25.33
4.1 10 .
eV
h XnmX X
c
eV
h X mX X
m
X
s
eV s
h X X
eV sX
h X eV s
X eV s
−
−
−
−
−
 =
 =
 =
 = =
=

4. 4
Re-arranging eq.(2) & eq.(3), as
0.65 420 420
420 0.65 420
420 0.65 420
420 ( 0.65 ) (4)
eVX nm hc X nm
X nm hc eVX nm
hc X nm eVX nm
hc nm eV




= −
 = −
 = +
 = + − − −
2 9
1.69
3.10 10 10
1.69 310 310
310 (1.69 ) (5)
hc
eV
X X m
hc eVX nm X nm
hc nm eV



−
= −
 = +
 = + − − −
Equating eq.(4) & eq.(5), we have
420 ( 0.65 ) 310 (1.69 )
420 (1.69 )
310 ( 0.65 )
1.3548( 0.65 ) (1.69 )
1.3548 0.88062 1.69 0
0.3548 0.80938 0
2.28
nm eV nm eV
nm eV
nm eV
eV eV
eV eV
eV
eV
 


 
 


+ = +
+
 =
+
 + = +
 + − − =
 − =
 =