Solveeit Logo
Login

Question

Question: The energy gap of silicon is \(1.14\,eV\). The maximum wavelength at which silicon starts energy abs...

The energy gap of silicon is 1.14 eV1.14\,eV. The maximum wavelength at which silicon starts energy absorption, will be (h=6.62×10−34Js,  c=3×108ms−1)\left( {h = 6.62 \times {{10}^{ - 34}}Js,\,\,c = 3 \times {{10}^8}m{s^{ - 1}}} \right)
A. 10.888A∘10.888\mathop A\limits^\circ
B. 108.88A∘108.88\mathop A\limits^\circ
C. 1088.8A∘1088.8\mathop A\limits^\circ
D. 10888A∘10888\mathop A\limits^\circ

Explanation

Solution

This is a direct problem where all the values are known to us, even the constant term is given. Now using the energy formula we can calculate the wavelength. As this energy has a relation between Plank’s constant speed of the silicon that is equals to speed of light and the wavelength. Putting and solving we can find the wavelength.

Formula used:
We know the energy formula,
E=hvλE = \dfrac{{hv}}{\lambda }
Where, energy gap of silicon is E=1.14eVE = 1.14eV, Planck constant is h=6.62×10−34Jsh = 6.62 \times {10^{ - 34}}Js and Velocity of the silicon is vv which equals the speed of the light is c=3×108ms−1c = 3 \times {10^8}m{s^{ - 1}}.

Complete step by step answer:
As per the problem we have the energy gap of silicon is 1.14eV1.14eV , h=6.62×10−34Jsh = 6.62 \times {10^{ - 34}}Js and c=3×108ms−1c = 3 \times {10^8}m{s^{ - 1}} .
We need to calculate the maximum wavelength at which silicon starts energy absorption.
We know the energy formula,
E=hvλE = \dfrac{{hv}}{\lambda }
Where, Energy gap of silicon is E=1.14eVE = 1.14eV.
The energy must be converted to Volt.
Hence, 1eV=1.6×10−19V1eV = 1.6 \times {10^{ - 19}}V
So, E=1.14×1.6×10−19VE = 1.14 \times 1.6 \times {10^{ - 19}}V
Planck constant is h=6.62×10−34Jsh = 6.62 \times {10^{ - 34}}Js.
Velocity of the silicon is vv which equals the speed of the light is c=3×108ms−1c = 3 \times {10^8}m{s^{ - 1}}.

Now putting the known values in the above energy formula we have,
1.14×1.6×10−19V=6.62×10−34Js×3×108ms−1λ1.14 \times 1.6 \times {10^{ - 19}}V = \dfrac{{6.62 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{\lambda }
Now rearranging the above equation we will get,
λ=6.62×10−34Js×3×108ms−11.14×1.6×10−19V\lambda = \dfrac{{6.62 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{{1.14 \times 1.6 \times {{10}^{ - 19}}V}}
On solving further we will get,
λ=6.62×3×10−34+8+191.14×1.6 ⇒λ=19.86×10−71.824m\lambda = \dfrac{{6.62 \times 3 \times {{10}^{ - 34 + 8 + 19}}}}{{1.14 \times 1.6}} \\\ \Rightarrow \lambda = \dfrac{{19.86 \times {{10}^{ - 7}}}}{{1.824}}m

Cancelling diving numerator and denominator by 1.8241.824 we will get,
λ=10.888×10−7m\lambda = 10.888 \times {10^{ - 7}}m
⇒λ=10888×10−10m\Rightarrow \lambda = 10888 \times {10^{ - 10}}m
We know,
1A∘=10−10m1\mathop A\limits^\circ = {10^{ - 10}}m
Hence λ=10888A∘\lambda = 10888\mathop A\limits^\circ .

Therefore the correct option is (D)\left( D \right).

Note: From the above energy wavelength relation we can conclude that the wavelength of a particle is inversely proportional to the energy absorption of the particle.Note that in shorter wavelengths the energy absorption is larger and vice versa.