Question

A semiconductor of energy band gap of 3eV is exposed to light. Calculate the wavelength of light emitted by it

Answer 1

Answer:

3 nm

Explanation:

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Answer 2

Answer:

The wavelength of light emitted by it is 4.125×10⁻⁷m.

Explanation:

We will use the following formula to solve this question,

$E=\frac{hc}{\lambda}$         (1)

Where,

E=energy in joule

h=planck's constant=6.6×10⁻³⁴J-second

c=speed of light in vacuum=3×10⁸m/s

λ=wavelength of light used

From the question we have,

The energy bandgap(E)=3eV

Now let's convert this energy(eV) into joule,

$E=3\times 1.6\times 10^-^1^9$

$E=4.8\times 10^-^1^9J$     (2)

By substituting the value of E,h, and c in equation (1) we get;

$4.8\times 10^-^1^9=\frac{6.6\times 10^-^3^4\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.6\times 10^-^3^4\times 3\times 10^8}{4.8\times 10^-^1^9}$

$\lambda=4.125\times 10^-^7m$

Hence, the wavelength of light emitted by it is 4.125×10⁻⁷m.