Answer:
0.5m^2/Vs and 0.14m^2/Vs
Explanation:
To calculate the mobility of electron and mobility of hole for gallium antimonide we have,
[tex]\sigma = n|e|\mu_e+p|e|\mu_h[/tex] (S)
Where
e= charge of electron
n= number of electrons
p= number of holes
[tex]\mu_e=[/tex] mobility of electron
[tex]\mu_h=[/tex]mobility of holes
[tex]\sigma =[/tex] electrical conductivity
Making the substitution in (S)
Mobility of electron
[tex]8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)[/tex]
[tex]0.639=\mu_e+\mu_h[/tex]
Mobility of hole in (S)
[tex]2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))[/tex]
[tex]0.1436 = 7.6*10^{-3}\mu_e+\mu_h[/tex]
Then, solving the equation:
[tex]0.639=\mu_e+\mu_h[/tex] (1)
[tex]0.1436 = 7.6*10^{-3}\mu_e+\mu_h[/tex] (2)
We have,
Mobility of electron [tex]\mu_e = 0.5m^2/V.s[/tex]
Mobility of hole is [tex]\mu_h = 0.14m^2/V.s[/tex]