In an n-type semiconductor, the Fermi level lies 0.3 eV below the conduction band at 300 K.
If the temperature is increased to 330 K, where does the new position of the Fermi level lie?

1. 0.55 eV below conducton band
2. 0.44 eV below conduction band
3. 0.33 eV below conduction band
4. 0.27 eV below conduction band

Answer is given as (d) 0.27eV ... pls explain how

Dear Student,

Please find below the solution to the asked query:

We have $E_c-E_f=KT\log \left(\frac{N_c}{N_d}\right)$
So at 300 K temperature we have
$$\begin{gathered} 0.3=300\times K\log \left(\frac{N_c}{N_d}\right) \\ Klog\left(\frac{N_c}{N_d}\right)=\frac{0.3}{300}=\frac{1}{1000} \end{gathered}$$
So at 330 degree temperature we have
$E_c-E_f=KT\log \left(\frac{N_c}{N_d}\right)=330\times \frac{1}{1000}=0.33 \text{eV}$
So Fermi level will be 0.33eV bellow conduction band. 
So answer should be 0.33 eV not 0.27eV
Hope this information will clear your doubts about this topic.

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