The acceptor level of a p-type semiconductor is 6 eV. The maximum wavelength of light which can create a hole would be (Given hc = 1242 eVnm.)

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Solution

Q.
The acceptor level of a p-type semiconductor is 6 eV. The maximum wavelength of light which can create a hole would be (Given hc = 1242 eVnm.) [2024]

1 407 nm

2 207 nm

3 103.5 nm

4 414 nm

Ans.
(2)

$Energy = \frac{h c}{λ};$

$E = \frac{1242}{λ (nm)} eV \Rightarrow 6 = \frac{1242}{λ (nm)}$

$\Rightarrow λ = \frac{1240}{6} = 207 nm$

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