1. When ultraviolet beams of wavelengths 80 nm and 110 nm struck lead, photoelectrons with maximum energies of 11.390 eV and 7.154 eV were produced. 2. Using the Planck constant equation and these values, the Planck constant h was calculated to be 6.63136×10-34 J·s. 3. The work function of lead was also calculated, found to be 4.11 eV for the 80 nm beam and 4.096 eV for the 110 nm beam.

1. 1
When two ultraviolet
beams of wavelengths
λ1=80 nm and λ2 = 110
nm fall on a lead
surface, they produce
photoelectrons with
maximum energies
11.390 eV and 7.154
eV, respectively. (a)
Estimate the numerical
value of the Planck
constant. (b) Calculate
the work function, the
cutoff frequency, and
the cutoff wavelength of
lead.

2. 2
Solution
We know that
(1)
where is the photon energy
in which is the Planck's
constant and is the frequency
of photon, is the work
function of the metal and is
the kinetic energ
hc
h W K
hc
K W
h
h
c
W
K






= = +
 = − − − −
=
y of the electron
leaving the metal.
1 2
1 2
2 1
1 2
1 2 1 2
1 2 1 2
2 1
For two ultraviolet beams
&
1 1
( ) ( )
(2)
hc hc
K W K W
K K hc hc
K K
h X
c
 
 
   
 
 
= − = −
−
 − = − =
−
 = − − −
−

3. 3
1 2
1 2
8
Using
80 , 110 ,
11.390 , 7.154 ,
2.998 10 , in eq.(2)
nm nm
K eV K eV
m
c X
s
 
= =
= =
=
8
2
8
8
17
15
11.390 7.154
110 80
110 80
2.998 10
4.236 8800
30 2.998 10
37276.8
89.94 10
414.46 10 .
4.1446 10 .
eV eV
h
nm nm
nmX nm
X
m
X
s
eV nm
X
m
nm X
s
eVnms
X m
X eV s
X eV s
−
−
−
=
−
=
=
=
=

4. 4
19
17
19
36
34
We know that
1 1.6 10
414.46 10
1.6 10 .
663.136 10 .
6.63136 10 .
eV X J
h X
X X J s
X J s
X J s
−
−
−
−
−
=
 =
=
=
1
1
2
2
34 8
9
17 19
2
( )
(3)
(4)
6.626 10 . 2.998 10
W
80 10
11.39
0.248 10 0.625 10 11.39
0.155 10 11.39 4.11
b
hc
W K
hc
W K
m
X J sX X
s
X m
eV
X X X eV eV
X eV eV eV


−
−
−
= − −−−
= − −−−
 =
−
= −
= − =

5. 5
34 8
9
17
17 19
,
6.626 10 . 2.998 10
W
110 10
7.154
0.180 10 7.154
0.180 10 0.625 10 7.154
11.25 7.154 4.096
Similarly
m
X J sX X
s
X m
eV
X J eV
X X X eV eV
eV eV eV
−
−
−
−
 =
−
= −
= −
= − =
0
19
0 34
15 15
15
We know that is called
the threshold or cutoff frequency
of the metal,
So,
4.11 1.6 10
6.626 10 .
0.99 10 1 10
10
W
h
W X X J
h X J s
X Hz X Hz
Hz


−
−
=
= =
= =
=

6. 6
8
0 15
0
7
9
,
2.998 10
10
2.998 10
299.8 10
299.8
300
Now
m
X
c s
Hz
X m
X m
nm
nm


−
−
= =
=
=
=
