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In Hall effect measurements a plate of width h=10mm and length l= 50mm made of p-type semiconductor was placed in a magnetic field with induction B= 5.0kG. A potential difference V= 10V was applied across the edges of the plate. In this case the hall field is V_(H)= 50mV and resistivity rho= 2.5 Omega, cm. Find the concentration of holes and hole mobility. |
Answer» Solution : We SHALL ignore minority CARRIES. Drifting holes experience a sideways force in the magnetic FIELD and react by setting up a Hall electric field `E_(y)` to conuterbalance it.Thus `v_(x)B=E_(y)=(V_(H))/(h)` If the concentration of carries is `n` then `j_(x)= n e v_(x)` Hence `n=(J_(x))/(ev_(x))=((j_(x))/(eV_(H)))/(hB)=(j_(x)hB)/(eV_(H))` Also using `j_(x)= sigmaE_(x)=E_(x)//rho=(V)/(rho l)` we get `n=(Vh B)/(e rhol V_(H))` Substituting the data (note that in `MKS` unit `B= 5.0kG= 0.5T)` `rho= 2.5xx10^(-2) ohm-m` we get `n=4.99xx10^(21)m^(-3)` `= 4.99xx10^(15)per cm^(3)` Also the mobility is` u_(0)=(v_(x))/(E_(x))=(V_(H))/(nB)xx(l)/(V)=(V_(H)l)/(hBV)` Substitution gives `u_(0)= 0.05m^(2)//V-s` |
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