Question

If the energy difference between the ground state and excited state of an atom is $4.4\times {{10}^{-19}}J$ . The wavelength of photon required to produce this transition is
(A) $3.5\times {{10}^{-7}}m$
(B) $4.5\times {{10}^{-7}}nm$
(C) $3.5\times {{10}^{-8}}m$
(D) $4.5\times {{10}^{-7}}m$

Answer 1

Hint: The ground state and the excited state are the energy levels of an atom with some energy differences between them. The electron after gaining energy shifts from ground state to excited state and vice versa.

Complete answer:
Let us understand the energy levels of an atom and their relationship with the photons;
The Bohr’s model of an atom describes the energy levels where electrons reside. The difference in energy levels emits the energy in the form of photons.
The energy emitted in the form of photons can be stated by the mathematical relationship i.e.
\[\begin{align}
& E=h\nu \\
& E=\dfrac{hc}{\lambda } \\
\end{align}\]
where,
E = energy of photon
c = speed of light = $3\times {{10}^{8}}m{{s}^{-1}}$
h = planck's constant = $6.626\times {{10}^{-34}}Js$
$\lambda $ = wavelength
Generally, a photon is a packet of energy travelling at the speed of light. Also, wavelength cannot be constant, it depends upon the energy emitted and how much a photon can carry.
The given illustration can be solved as,
Given data,
Energy difference = $\Delta $E = $4.4\times {{10}^{-19}}J$
Thus, wavelength can be determined as,
$\lambda =\dfrac{hc}{\Delta E}$
Putting the values, we get,
$\begin{align}
& \lambda =\dfrac{6.626\times {{10}^{-34}}Js\times 3\times {{10}^{8}}m{{s}^{-1}}}{4.4\times {{10}^{-19}}J} \\
& \lambda =4.517\times {{10}^{-7}}m \\
\end{align}$

Therefore, option (D) is correct.

Note: Students may get confused between the option (B) and (D). Do note to analyse units properly before you solve the illustration. The wavelength of photon is analysed to be $4.517\times {{10}^{-7}}m=451.7\times {{10}^{-9}}m=451.7nm$.