\(=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9} \times 1.6 \times 10^{-19}} eV\)

\(=\frac{6.63 \times 3 \times 100}{248 \times 1.6}\)

\(=0.05 eV \times 100=5 eV\)

Now using \(E =\phi+ K . E\)

\(5=3+ K . E\)

\(K . E .=2 eV =3.2 \times 10^{-19} J\)

for debroglie wavelength \(\lambda=\frac{ h }{ mv }\)

\(K . E =\frac{1}{2} mv ^{2}\)

so \(v =\sqrt{\frac{2 KE }{ m }}\)

hence \(\lambda=\frac{ h }{\sqrt{2 KE \times m }}\)

\(=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 3.2 \times 10^{-19} \times 9.1 \times 10^{-31}}}\)

\(=\frac{6.63}{7.6} \times \frac{10^{-34}}{10^{-25}}=\frac{66.3 \times 10^{-10} m }{7.6}\)

\(=8.72 \times 10^{-10} m\)

\(\approx 9 \times 10^{-10} m\)

\(=9 ^o\)