Question

Light emitted during the deexcitation of electron from $n=3$ to $n=2$, when incident on a metal, photoelectrons are just emitted from that metal. In which of the following deexcitations photoelectric effect is not possible?

• A. From $n=2$ to $n=1$
• B. From $n=3$ to $n=1$
• C. From $n=5$ to $n=2$
• D. From $n=4$ to $n=3$

Answer 1

The correct option is D From $n=4$ to $n=3$

Let the energy of electron in the first orbit be $E_o$

Then by Bohr's model, energy of electron in $n^{th}$ orbit is $E_n=\frac{E_o}{n^2}$

Energy of photon released in deexcitation of photon from $n_i^{th}$ orbit to $n_f^{th}$ orbit is given by:

$\Delta E=E_o(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

It is given that photon emitted due to electron transition from $n=3$ to $n=2$ is just enough for emission of photoelectron from the metal. Hence, minimum energy of photon required for emission is:

$E_{min}=E_o(\frac{1}{4}-\frac{1}{9})=\frac{5E_o}{36}\approx 0.14E_o$

Option A: Energy of released photon is $\Delta E=E_o(\frac{1}{1}-\frac{1}{4})=0.75E_o>E_{min}$

Option B: Energy of released photon is $\Delta E=E_o(\frac{1}{1}-\frac{1}{9})=0.89E_o>E_{min}$

Option C: Energy of released photon is $\Delta E=E_o(\frac{1}{4}-\frac{1}{25})=0.21E_o>E_{min}$

Option D: Energy of released photon is $\Delta E=E_o(\frac{1}{9}-\frac{1}{16})=0.05E_o<E_{min}$

Hence, only in option D, emission of photoelectron cannot take place.