Correct Answer  Option 4 : 0.173 V
Concept:
The Builtin Potential of a pn junction diode is given by:
\({V_o} = \frac{{kT}}{q}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)
Where NA = doping density of holes
ND = doping density of electrons
Also, the law of massAction states that:
\(n_i^2 = {N_A}{N_D}\)
Calculation:
\(n + \)
\({10^{18}}/c{m^3}\)

\(n\)
\({10^{15}}/c{m^3}\)

n^{+} = 10^{18}/cm^{3}
n = 10^{15}/cm^{3}
n_{i} = 10^{10}/cm^{3}
The n^{+} Junction here has more electrons than the n Junction [Analogous to pn Junction]
Hole concertation in n – region using mass action law will be:
\(p= \frac{{n_i^2}}{N} = \frac{{{{10}^{20}}}}{{{{10}^{15}}}}\) = N_{A}
Putting on the respective values, we get:
\({V_o} = \frac{{KT}}{2}{l_n}\left( {\frac{{N_D^ + + {N_A}}}{{n_i^2}}} \right) = 0.173V\)