A potential barrier of 0 . 4V exists across a p-n junction. An...

Applying conservation of energy

$- e{V}_{n} + \dfrac{1}{2}m{v}_{i}^{2} = - e{V}_{p} + \dfrac{1}{2}m{v}_{f}^{2}$

We know, ${V}_{n} - {V}_{p} =$potential barrier$= 0 . 4V$

Then, we have, $- e\left({V}_{n} - {V}_{p}\right) + \dfrac{1}{2}m{v}_{i}^{2} = \dfrac{1}{2}m{v}_{f}^{2}$

$⇒ - 1 . 6 \times {10}^{- 19} \times 0 . 4 + \dfrac{1}{2} \times 9 \times {10}^{- 31} \times 36 \times {10}^{10} = \dfrac{1}{2} \times 9 \times {10}^{- 31}{v}_{f}^{2}$

$⇒ - 0 . 64 \times {10}^{- 19} + 1 . 62 \times {10}^{- 19} = \dfrac{1}{2} \times 9 \times {10}^{- 31}{v}_{f}^{2}$

$⇒ 0 . 98 \times {10}^{- 19} = 4 . 5 \times {10}^{- 31}{v}_{f}^{2}$

$⇒ {v}_{f}^{2} = \dfrac{0 . 98 \times {10}^{12}}{4 . 5}$

$⇒ {v}_{f} = 0 . 466666 \times {10}^{6}$

$= 4 . 6666 \times {10}^{5}m{s}^{- 1} = \dfrac{14}{3} \times {10}^{5}m{s}^{- 1}$