Question

The energy, II and III energy levels of a certain atom are E,$\dfrac{{4E}}{3}$ and 2E respectively. A photon of wavelength $\lambda $ is emitted during a transition from III to I. What will be the wavelength of emission for transition II to I?
A.$\dfrac{\lambda }{2}$
B.$\lambda $
C. $2\lambda $
D.$3\lambda $

Answer 1

Hint: During transition of electron from higher energy level to lower energy level, energy is released in the form of photons having energy equal to the energy difference between the two energy levels.
Formula used: $\dfrac{{hc}}{\lambda } = E$

Complete step by step answer:
Calculate energy of photon released on transition from 3$ \to $1 energy level
Energy of photon released=
 $\begin{gathered}
   = {E_3} - {E_1} \\
   = 2E - E \\
   = E \\
\end{gathered} $
Hence, by the formula
$\dfrac{{hc}}{\lambda } = E$
Calculate energy of photons released on transition from 2$ \to $1 energy level.
${E_2} - {E_1} = \dfrac{{4E}}{3} - E = \dfrac{{4E - 3E}}{3} = \dfrac{E}{3}$
Find the wavelength for transition from 2$ \to $1 energy level
$\dfrac{{hc}}{\lambda } = \dfrac{E}{3}$
$\lambda = \dfrac{{3hc}}{E}$
Put $\dfrac{{hc}}{E} = \lambda $
$\lambda ' = 3\lambda $

Option D is correct.

Additional information:
According to plank, energy is in the form of small packets called quantum(plural- quanta). This one quantum is called a photon. The concept of photons and quanta comes from quantum mechanics and quantum theory. Photons always move with the speed of light. Photons have no mass but they have energy. The energy of each photon is inversely proportional to wavelength of the associated photon.

Note:
 During transition of electron from higher energy level to lower energy level, energy is released while during transition of electron from lower energy level to higher energy level, energy is absorbed.