Solution:
(a) The energy of black body radiation is
E=2//- dnpdnq AW
(2nh)n ehwl2nkT - 1For the radiation we have p = Aw/c, so
where x = Aw/kT. Hence Q = n + 1.
(b) The Debye Model regards solid as an isotropic continuous medium
with partition function
nN nN[ i=i 1 j=1
Z(T,V) = exp -AEw;/2kT n[l- exp(-f~wi/kT)]-l
The Holmholtz free energy isnN
h nN
F=-kTlnZ= -~w;+kT~ln[l-exp(-hw;/kT)].
i=l i= 1
2When N is very large,
nN
-+ $ lwD wn-'dw ,
i=lwhere WD is the Debye frequency. So we have
AWD + (kT)n+i ___ xn-' ln[l- exp(-x)]dx ,
(ftwD)n n2N lXDn2N
2(n + 1)F= -
where XD = hwD/kT. Hence
cv=-T(~)mTn, a2 F
i.e., B = n.