Question

A photon and an electron both have wavelength 1 Angstrom. The ratio of energy of photon to that of electron is:
$\begin{align}
  & (a)1 \\
 & (b)0.012 \\
 & (c)82.35 \\
 & (d){{10}^{-10}} \\
\end{align}$

Answer 1

Hint: We will use the energy equation for both photon and electron to find their respective energy. We can then divide them to easily get the ratio of their energy. We should be using Planck’s equation to find the energy of the photon as it is massless and a combination of Planck’s equation and Einstein’s equation to calculate the energy of the electron.

Complete answer:
Since the wavelength of photon and electron is given to be equal, let it be termed by $\lambda $.
Now, let the energy of the photon be ${{E}_{p}}$ . This can be calculated using Planck’s equation which is given by:
$\Rightarrow E=\dfrac{hc}{\lambda }$
Where,
$h$ is the Planck’s constant whose value is equal to $6.626\times {{10}^{-34}}{{m}^{2}}kg{{s}^{-1}}$
$c$ is the speed of light whose value is $3\times {{10}^{8}}m{{s}^{-1}}$
$\lambda $ is the wavelength of the photon which is given to be 1 Angstrom .
Now, we have:
$\Rightarrow {{E}_{p}}=\dfrac{hc}{\lambda }$ [Let this expression be equation number (1)]
For an electron, we have:
Firstly, from the Planck’s equation:
$\Rightarrow {{E}_{e}}=\dfrac{hc}{\lambda }$
Now, equating it with Einstein’s equation. We get:
$\begin{align}
  & \Rightarrow {{m}_{e}}{{c}^{2}}=\dfrac{hc}{\lambda } \\
 & \Rightarrow c=\dfrac{h}{\lambda {{m}_{e}}} \\
\end{align}$
Using this value of $c$to find energy of electron, we get:
$\Rightarrow {{E}_{e}}=\dfrac{1}{2}{{m}_{e}}{{\left( \dfrac{h}{\lambda {{m}_{e}}} \right)}^{2}}$
On simplifying, we get:
$\Rightarrow {{E}_{e}}=\dfrac{{{h}^{2}}}{2{{m}_{e}}{{\lambda }^{2}}}$ [Let this expression be equation number (2)]
Now, the ratio of energy of photon to that of electron is:
$\begin{align}
  & \Rightarrow \dfrac{{{E}_{p}}}{{{E}_{e}}}=\dfrac{\dfrac{hc}{\lambda }}{\dfrac{{{h}^{2}}}{2{{m}_{e}}{{\lambda }^{2}}}} \\
 & \Rightarrow \dfrac{{{E}_{p}}}{{{E}_{e}}}=\dfrac{2{{m}_{e}}\lambda c}{h} \\
\end{align}$
Now, putting all the values we get:
 $\begin{align}
  & \Rightarrow \dfrac{{{E}_{p}}}{{{E}_{e}}}=\dfrac{2\times 9.1\times {{10}^{-31}}\times {{10}^{-10}}}{6.626\times {{10}^{-34}}} \\
 & \Rightarrow \dfrac{{{E}_{p}}}{{{E}_{e}}}=82.35 \\
\end{align}$
Hence, the ratio of energy of photon to that of electron comes out to be $82.35$ .

Hence, option (c) is the correct option.

Note:
Equating the energy using Planck’s equation would give us the wrong solution because it is used to measure the energy of photons only. And since photons are massless, Einstein’s energy equations are invalid for photons. Thus, we needed to bring them under the same parameters to calculate the ratio of their energies.