Given data:

• $T=300 K$
• $k=1.38\times10^{-23} J/K$
• $E_F= 5.5 eV$

• $f(E)=0.9$

   

To find: E = ?

Solution:

$kT=\dfrac{1.38 \times10^{-23} \times300 }{{1.6 \times10^{-19}}}$

$\therefore kT=0.0259$

$f(E)=\dfrac { 1}{ 1+e^\frac{(E-E_f)}{kT}}$

$0.9=\dfrac { 1}{ 1+e^\frac{(E-5.5)}{0.0259}}$

${ 1+e^\dfrac{(E-5.5)}{0.0259}}=\dfrac{1}{0.9}$

${ 1+e^\dfrac{(E-5.5)}{0.0259}}=1.1111$

$\therefore e^\dfrac{(E-5.5)}{ 0.0259}={ 1.1111}-1$

$\therefore e^\dfrac{(E-5.5)}{ 0.0259}={ 0.1111}$

Taking log on both sides,

$\log {e^{(E-5.5)\over 0.0259}}=\log{ 0.1111}$

$\therefore\dfrac{(E-5.5)}{ 0.0259}=-0.9543$

${{(E-5.5) }}=-0.9543\times0.0259$

${{(E-5.5) }}=-0.0247$

$E=-0.0247+5.5$

$E=5.4753$

$\therefore E=5.4753 \ eV$