# 3-2 Assume that your skin can be considered a blackbody. One... C = 308 K to find displacement law, ```3-2
Assume that your skin can be considered a blackbody. One can then use Wien’s
displacement law, &quot; max T = 0.289 8 # 10
&quot; max =
3-4
(a)
\$2
0
m % K with T = 35 C = 308 K to find
0.289 8 # 10 \$2 m % K
= 9.41 # 10 \$6 m = 9 410 nm .
308 K
From Stefan’s law, one has
P
4
= &quot; T . Therefore,
A
P
4
#8
2 4
6
2
= 5.7 &quot; 10 W m K (3000 K ) = 4.62 &quot; 10 W m .
A
(
3-5
)
P
75 W
2
(b)
A=
(a)
Planck’s radiation energy density law as a function of wavelength and
&quot;u
8# hc
= 0 and setting
temperature is given by u (&quot; , T ) = 5 hc &quot; T
. Using
&quot;#
&quot; e B \$1
4.62 &quot; 10 6 W m2
=
4.62 &quot; 106 W m2
= 16.2 mm .
(
)
hc
x=
, yields an extremum in u (&quot; , T ) with respect to λ. The result is
&quot; max kB T
\$
&quot;1
hc ' hc #max kBT hc # max kBT
&quot;x
0 = &quot;5 + &amp;
e
e
&quot;1
or x = 5 1 &quot; e .
)
% # max k BT (
(
)
(
)
(b)
Solving for x by successive approximations, gives x &quot; 4.965 or
# hc &amp;
&quot; max T = % ( (4.965) = 2.90 ) 10 *3 m +K .
\$ kB '
(a)
In general, L =
!
3-7
)(
n&quot;
where n = 1, 2 , 3, K defines a mode or standing wave
2
pattern with a given wavelength. As we wish to find the number of possible
2L
values of n between 2.0 and 2.1 cm, we use n =
&quot;
200
= 200
2.0
200
n (2.1 cm ) = (2)
= 190
2.1
&quot; n = 10
n( 2.0 cm) = (2)
As n changes by one for each allowed standing wave, there are 10 standing
waves of different wavelength between 2.0 and 2.1 cm.
!
!
2m
!
!
(b)
The number of modes per unit wavelength per unit length is
&quot;n
10
(200) = 0.5 cm \$2 .
=
L&quot; # 0.1
(c)
For short wavelengths n is almost a continuous function of λ. Thus one may use
&quot; n \$ 1 ' \$ dn '
2L dn 2L
= &amp; ) &amp; ) . As n =
calculus to approximate
,
and
=
%
(
%
(
L&quot; #
L d*
&quot; d&quot; &quot;2
&quot; 1 %&quot;
\$# '&amp;\$#
L
&quot; 1 %&quot;
\$# '&amp;\$#
L
(d)
dn % 2
. This gives approximately the same result as that found in (b):
'=
d( &amp; (2
dn % 2
2
= 0.5 cm )2 .
'&amp; = 2 =
d(
(
(2.0 cm)2
For short wavelengths n is almost a continuous function of λ, n =
function.
2L
is a discrete
&quot;
The energy per photon, E = hf and the total energy E transmitted in a time t is Pt where
power P = 100 kW . Since E = nhf where n is the total number of photons transmitted in
3-10
(
5
)
the time t, and f = 94 MHz , there results nhf = (100 kW)t = 10 W t , or
5
5
10 J s
n 10 W
=
=
94 &quot; 106 s#1 = 1.60 &quot; 10 30 photons s .
#34
t
hf
6.63 &quot; 10
Js
3-14
!
(
)
1 240 eV nm
hc
=
&quot; 2.24 eV = 1.30 eV
\$ &quot;#
350 nm
(a)
K = hf &quot; # =
(b)
At &quot; = &quot;c , K = 0 and &quot; =
(a)
&quot;=
hc 1 240 eV nm
=
= 554 nm
#
2.24 eV
!
3-16
1 240 eV nm
hc
\$ K, &quot; =
# 2.23 eV = 1.90 eV
#
300 nm
1 240 eV nm
&quot;1.90 eV e = 1.20 V
400 nm e
The energy of one photon of light of wavelength &quot; = 300 nm is
(b)
3-17
Vs =
!
E=
(a)
hc 1 240 eV nm
=
= 4.13 eV .
&quot;
300 nm
As lithium and beryllium have work functions that are less than 4.13 eV, they
will exhibit the photoelectric effect for incident light with this energy. However,
mercury will not because its work function is greater than 4.13 eV.
(b)
The maximum kinetic energy is given by K =
hc
, so
&quot; #\$
1 240 eV nm
&quot; 2.3 eV = 1.83 eV , and
300 nm
1 240 eV nm
K ( Be) =
&quot; 3.9 eV = 0.23 eV .
300 nm
K (Li) =
3-18
3-20
!
(a)
K max = eVs = s(0.45 V) = 0.45 eV
(b)
&quot;=
(c)
&quot;c =
1 240 eV nm
hc
=
\$ 0.45 eV = 2.03 eV
# \$K
500 nm
hc 1 240 eV nm
=
= 612 nm
#
2.03 eV
hc
hc
&quot;# % # =
&quot; K max ;
\$
\$
hc
First Source: &quot; =
\$ 1.00 eV .
#
hc
2hc
\$ 4.00 eV .
Second Source: &quot; = # \$ 4.00 eV =
#
2
As the work function is the same for both sources (a property of the metal),
hc
2 hc
hc
# 100 eV =
# 4.00 eV \$
= 3.00 eV and
&quot;
&quot;
&quot;
hc
&quot;=
\$ 1.00 eV = 3.00 eV \$1.00 eV = 2.00 eV .
#
K max = hf &quot; # =
```