Question

The following diagram indicates the energy levels of certain particles when the system moves from 4E to E. A photon of wavelength $\lambda_1$ is emitted. The wavelength of photon produced during its transition from $\dfrac{7}{3}E$ level to E is $\lambda_2$. Then the ratio $\dfrac{\lambda_1}{\lambda_2}$ will be

Answer 1

We know that photoelectric effect is the process observed when the incident radiation of light having energy more than the threshold energy or the work function is incident surface, emits charged particles, called the photons. Using the formula of photoelectric effect we can solve the following questions.

Formula used:
$E=hv=\dfrac{hc}{\lambda}$

Complete step-by-step solution:
We know that light exhibits dual nature, that is it can emit photons, and also shows properties of waves like in the case of electromagnetic spectrum. This property of the light called the dual nature was studied and there were two cases called the photoelectric effect of light and the other as the reverse photoelectric effect.
We know from the dual nature of light, that it is both wave and also contains particles or photons. Photoelectric electric effect is a foundation to prove that light has particle nature. As defined above, it was found that light particles are nothing but species called photons. From Planck’s equation, we know the energy $E$ of light as a wave can be expressed as $E=hv=\dfrac{hc}{\lambda}$, where $h$ is the Planck’s constant and $v$ denotes the frequency of light wave with speed $c$ and wavelength $\lambda$
From the diagram, we have
$4E-E=\dfrac{hc}{\lambda_1}$
$\implies 3E=\dfrac{hc}{\lambda_1}$
Similar $\dfrac{7E}{3}-E=\dfrac{hc}{\lambda_2}$
$\implies \dfrac{4E}{3}=\dfrac{hc}{\lambda_2}$
Taking the ratio, we have
$\implies \dfrac{3E}{\left(\dfrac{4E}{3}\right)}=\dfrac{\left(\dfrac{hc}{\lambda_1}\right)}{\left(\dfrac{hc}{\lambda_2}\right)}$
$\therefore \dfrac{\lambda_1}{\lambda_2}=\dfrac{9}{4}$
Thus the required ratio is $\dfrac{\lambda_1}{\lambda_2}=\dfrac{9}{4}$

Note: Here, it is important to note that the photon is nothing but the hydrogen atom which is undergoing excitation; hence we must compare the energy of the photon to the difference of the energy levels of the hydrogen. To solve such questions, always find the energy difference between the levels.