Question

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

Answer 1

Hint: 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,
$E = \dfrac{{hv}}{\lambda }$
Where, energy gap of silicon is $E = 1.14eV$, Planck constant is $h = 6.62 \times {10^{ - 34}}Js$ and Velocity of the silicon is $v$ which equals the speed of the light is $c = 3 \times {10^8}m{s^{ - 1}}$.

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

Now putting the known values in the above energy formula we have,
$1.14 \times 1.6 \times {10^{ - 19}}V = \dfrac{{6.62 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{\lambda }$
Now rearranging the above equation we will get,
$\lambda = \dfrac{{6.62 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{{1.14 \times 1.6 \times {{10}^{ - 19}}V}}$
On solving further we will get,
$\lambda = \dfrac{{6.62 \times 3 \times {{10}^{ - 34 + 8 + 19}}}}{{1.14 \times 1.6}} \\
\Rightarrow \lambda = \dfrac{{19.86 \times {{10}^{ - 7}}}}{{1.824}}m$

Cancelling diving numerator and denominator by $1.824$ we will get,
$\lambda = 10.888 \times {10^{ - 7}}m$
$ \Rightarrow \lambda = 10888 \times {10^{ - 10}}m$
We know,
$1\mathop A\limits^\circ = {10^{ - 10}}m$
Hence $\lambda = 10888\mathop A\limits^\circ $.

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

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.