In [1]:

```
import math
# Given
E_FO = 7.0 # in eV
e = 1.6*10**-19 # in coulombs
E_FO *= e # in Joules
me = 9.1*10**-31 # in Kg
# Calculations and Results
v_f = math.sqrt(2*E_FO/me)
print("Speed of the conduction electrons in m/s is {0:.4g}".format(v_f))
```

In [2]:

```
# Given
e = 1.6 * 10 ** -19 # in coulombs
Eg = 1.1 # in eV
Eg *= e # in Joules
h = 6.6 * 10 ** -34 # in Js
c = 3 * 10 ** 8 # in m/s
# Calculations and Results
lambda_val = h * c / Eg
print("Wavelength of light that can be absorbed by an Si photo-detector"
" at Eg=1.1 eV in micro meter is {0:.4g}".format(lambda_val * 10 ** 6))
print("Hence the light of wavelength 1.31 micro meter and 1.55 micro meter will not "
"be absorbed by Si and thus cannot be detected by detector")
```

In [3]:

```
import math
# Given
e = 1.6*10**-19 # in coulombs
h = 6.626*10**-34 # in Js
me = 9.1*10**-31 # in Kg
# let x=k*T
x = 0.026 # in eV
E = 5.0 # in ev
E *= e # in Joules
# Calculations and Results
g_E = (8*math.pi*math.sqrt(2))*(me/h**2)**(3./2)*math.sqrt(E) # in J**-1*m**-3
# conversion of units
g_E = g_E*10**-6*e # in eV**-1 cm**-3
print("density of states at the center of the band in cm**-3*J**-1 is {0:.4g}".format(g_E))
# part(b)
n_E = g_E*x # in cm**-3
print("No.of states per unit volume within kT about the center in cm**-3 is {0:.4g}".format(n_E))
# part(c)
# Given
E = 0.026 # in eV
E *= e # in joules
# Calculations and Results
g_E = (8*math.pi*math.sqrt(2))*(me/h**2)**(3./2)*math.sqrt(E) # in J**-1*m**-3
# conversion of units
g_E = g_E*10**-6*e # in eV**-1 cm**-3
print("density of states at at kT above the band in cm**-3*J**-1 is {0:.4g}".format(g_E))
# part(d)
n_E = g_E*x # in cm**-3
print(" No.of states per unit volume within kT about the center in cm**-3 is {0:.4g}".format(n_E))
# solved using the values taken from the solution of textbook
```

In [4]:

```
import math
# Given
e = 1.6*10**-19 # in coulombs
h = 6.626*10**-34 # in Js
me = 9.1*10**-31 # in Kg
d = 10.5 # in g/cm
Mat = 107.9 # g/mol
NA = 6.023*10**23 # mol**-1
E_ctr = 5.0 # in ev
E_ctr *= e # in Joules
# Calculations and Results
S_band = 2*(16*math.pi*math.sqrt(2)/3)*(me/h**2)**(3./2)*E_ctr**(3./2) # in states m**-3
# conversion of units
S_band *= 10**-6 # in states cm**-3
print("No. of states in the band in states cm**-3 is {0:.4g}".format(S_band))
n_Ag = d*NA/Mat
print("No.of atoms per unit volume in silver in atoms per cm3 is {0:.4g}".format(n_Ag))
```

In [5]:

```
import math
# Given
e = 1.6*10**-19 # in coulombs
h = 6.626*10**-34 # in Js
me = 9.1*10**-31 # in Kg
d = 8.96 # in g/cm
Mat = 63.5 # g/ mol
NA = 6.023*10**23 # mol**-1
n = d*NA/Mat # in cm**-3
n *= 10**6 # in m**-3
# Calculations and Results
E_FO = (h**2/(8*me))*(3*n/math.pi)**(2./3) # in J
E_FO /= e # in eV
print("Fermi energy at 0 Kelvin in eV is {0:.4f}".format(E_FO))
E_FO = (h**2/(8*me))*(3*n/math.pi)**(2./3) # in J
v_e = math.sqrt(6*E_FO/(5*me))
print("Average speed of conduction electrons in m/s is {0:.4g}".format(v_e))
```

In [6]:

```
# Given
e = 1.6*10**-19 # in coulombs
me = 9.1*10**-31 # in Kg
u_d = 43*10**-4 # in cm2/V/s
v_e = 1.22*10**6 # in m/s
# Calculations and Results
T = u_d*me/e
l_e = v_e*T
print("Mean free path of electrons in meters is {0:.4g}".format(l_e))
```

In [7]:

```
import math
# Given
e = 1.6*10**-19 # in coulombs
T = 373.0 # in kelvin
To = 273.0 # in kelvin
k = 1.38*10**-23 # in m2 kg /k/s2
# from table 4.3
E_FAO = 11.6 # in eV
E_FAO *= e # in J
x_A = 2.78
E_FBO = 7.01 # in eV
E_FBO *= e # in J
x_B = -1.79
# Calculations and Results
# Mott jones Equation
V_AB = (-math.pi**2*k**2/(6*e))*((x_A/E_FAO)-(x_B/E_FBO))*(T**2-To**2)
print("EMF in micro volts available from Al and Cu thermocouple with the given respective"
" temperatures at the junctions is {0:.4f}".format(V_AB*10**6))
```

In [8]:

```
import math
# Given
phi = 2.6 # in eV
e = 1.6*10**-19 # in coulombs
phi *= e # in Joules
Be = 3*10**4 # schottky coefficient in A/m2/K2
T = 1600.0 # in degree celsius
T += 273.0 # in Kelvin
k = 1.38*10**-23 # m2 kg s-2 K-1
d = 2*10**-3 # in m
l = 4*10**-2 # in in m
# Calculations and Results
# Richardson-Dushman Equation
J = Be*T**2*math.exp(-phi/(k*T))
A = math.pi*d*l
I = J*A
print("Saturation current in Amperes if the tube is operated at 1873 kelvin is {0:.4g}".format(I))
```