To solve this problem, we need to follow these steps:

1. Calculate the energy of the emitted photon using the photoelectric effect equation.

Use the energy levels of the hydrogen atom to find the initial state that corresponds to the emitted photon's energy.

The work function of the metallic electrode is $$0.5 \mathrm{~eV}$$, and the stopping voltage is $$0.47 \mathrm{~V}$$. We can use the equation for the photoelectric effect:

$$ E_{\text{photon}} = \text{Work function} + \text{Kinetic energy of ejected electron} $$

The kinetic energy of the ejected electron in electron volts (eV) is equal to the stopping voltage, which is $$0.47 \mathrm{~eV}$$. Therefore:

$$ E_{\text{photon}} = 0.5 \mathrm{~eV} + 0.47 \mathrm{~eV} = 0.97 \mathrm{~eV} $$

Now, we need to find the energy levels of the hydrogen atom. The energy level of an electron in the nth state of a hydrogen atom is given by:

$$ E_n = -13.6 \frac{1}{n^2} \mathrm{~eV} $$

The energy difference between the two states $$n$$ and the second excited state ($$n = 3$$) is given by the equation:

$$ \Delta E = E_n - E_3 = -13.6 \left( \frac{1}{n^2} - \frac{1}{3^2} \right) \mathrm{~eV} $$

We know that this energy difference must be equal to the energy of the photon:

$$ 0.97 = -13.6 \left( \frac{1}{n^2} - \frac{1}{9} \right) $$

To simplify, we multiply both sides by -1:

$$ -0.97 = 13.6 \left( \frac{1}{n^2} - \frac{1}{9} \right) $$

Next, we divide both sides by 13.6:

$$ -\frac{0.97}{13.6} = \frac{1}{n^2} - \frac{1}{9} $$

Which simplifies to:

$$ -\frac{0.97}{13.6} = \frac{1}{n^2} - \frac{1}{9} \Rightarrow -0.07132 = \frac{1}{n^2} - 0.1111 $$

Now, isolate $$\frac{1}{n^2}$$:

$$ \frac{1}{n^2} = -0.07132 + 0.1111 = 0.03978 $$

Thus:

$$ n^2 = \frac{1}{0.03978} \approx 25.14 $$

Therefore:

$$ n \approx \sqrt{25.14} \approx 5 $$

So the quantum number of the state '$$n$$' is approximately 5.

The correct option is Option A: 5.

Dual Nature Of Radiation

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