Question

What is the wavelength of a photon of energy 1eV?
$\begin{align}
  & \text{A}.12.4\times {{10}^{3}}\overset{o}{\mathop{A}}\, \\
 & \text{B}.2.4\times {{10}^{3}}\overset{o}{\mathop{A}}\, \\
 & \text{C}.0.4\times {{10}^{2}}\overset{o}{\mathop{A}}\, \\
 & \text{D}.1000\overset{o}{\mathop{A}}\, \\
\end{align}$

Answer 1

Hint: A photon’s energy depends on its frequency. In this question we have to find the wavelength of a photon, whose energy is given. We know that energy of a photon is inversely proportional to its wavelength. Here, to find the wavelength we use Planck’s relation.

Formula Used:
$E=h\nu $

Complete step-by-step answer:
In the question we are given energy of a photon
${{E}_{photon}}=1eV$
We are asked to find the wavelength of this photon.
Planck’s relation gives the equation for energy of a photon.
$E=h\nu $, where ‘E’ is energy of photon, ‘h’ is Planck’s constant, ‘$\nu $ ‘ is frequency.
The relation between frequency$\left( \nu \right)$ and wavelength$\left( \lambda \right)$ is given by
$\nu =\dfrac{c}{\lambda }$
By substituting for$\nu $ in Planck’s relation equation, we get
$E=\dfrac{hc}{\lambda }$ , where ‘c’ is speed of light and ‘$\lambda $’ is the wavelength.
From this equation, to find wavelength we can rewrite the equation as
$\lambda =\dfrac{hc}{E}$
Value of speed of light is,$c=3\times {{10}^{8}}$
Value of Planck’s constant is,$h=6.626\times {{10}^{-34}}$
Energy of photon is given in electron volts, $E=1eV$
Therefore energy of photon in Joules is, $E=1.602\times {{10}^{-19}}$
By substituting the above values in the equation, we get
$\lambda =\left( \dfrac{\left( 6.626\times {{10}^{-34}} \right)\left( 3\times {{10}^{8}} \right)}{\left( 1.602\times {{10}^{-19}} \right)} \right)$
By solving this we get the answer in meters.
We have to find the solution in $\overset{o}{\mathop{A}}\,$, for that we need to multiply the solution with ${{10}^{10}}$.
Therefore,
$\lambda =\left( \dfrac{\left( 6.626\times {{10}^{-34}} \right)\left( 3\times {{10}^{8}} \right)\left( {{10}^{10}} \right)}{\left( 1.602\times {{10}^{-19}} \right)} \right)$
By solving this, we get
$\lambda =12.408\times {{10}^{3}}\approx 12.4\times {{10}^{3}}$
Thus the wavelength of a photon of energy 1eV is $12.4\times {{10}^{3}}$.
Hence the correct answer is option A.

Note: Photon is simply the smallest discrete quantum of electromagnetic radiation. It is a mass less particle. Photons are the basic unit of light.
Planck’s relation says that energy of a photon is directly proportional to its frequency by a constant factor. This constant is called Planck’s constant (h) it’s value is $6.626\times {{10}^{-34}}$.
To convert meter (m) to angstrom $\left( \overset{o}{\mathop{A}}\, \right)$ we multiply meter with ${{10}^{10}}$.
$m\times {{10}^{10}}=\overset{o}{\mathop{A}}\,$.