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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
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The width of the depletion region of the $p-n$ junction is
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Verified Answer
The correct answer is:
$1000 \mathrm{~nm}$
$$
\Rightarrow \quad \beta=34
$$
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, $\quad 2 \times 10^5 \times d=0 \cdot 2$
$$
\Rightarrow \quad d=10^{-6} \mathrm{~m} \text { or } 1000 \mathrm{~nm}
$$
\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}
$$
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