Q159 / 160 · TS EAMCET 2018 (04 May Shift 2) · Physics

In a $p-n$ junction diode, an electric field of magnitude $2 \times 10^5 \mathrm{~V} /$ mexists in the depletion region. A particle with charge $-3 e$ can diffuse from $n$-side to $p$-side, if it has minimum kinetic energy $0.6 \mathrm{eV}$. The width of the depletion region of the $p-n$ junction is

  • A
    $300 \mathrm{~nm}$
  • B
    $600 \mathrm{~nm}$
  • Correct answer
    C
    $1000 \mathrm{~nm}$
  • D
    $1200 \mathrm{~nm}$

Solution

$$ \Rightarrow \quad \beta=34 $$ For depletion width $d$, developed potential is $$ V=E \cdot d=2 \times 10^5 \times d $$ For charged particle, experienced potential is $$ V=\frac{\text { Energy }}{\text { Charge }}=\frac{0 \cdot 6 \mathrm{eV}}{3 e}=0 \cdot 2 \mathrm{~V} $$ So, $\quad 2 \times 10^5 \times d=0 \cdot 2$ $$ \Rightarrow \quad d=10^{-6} \mathrm{~m} \text { or } 1000 \mathrm{~nm} $$

Hence, the correct answer is option: 'C'.

Relevant Questions