A Hydrogen-like atom has atomic number $$Z$$. Photons emitted in the electronic transitions from level $$n = 4$$ to level $$n = 3$$ in these atoms are used to perform photoelectric effect experiment on a target metal. The maximum kinetic energy of the photoelectrons generated is 1.95 eV. If the photoelectric threshold wavelength for the target metal is 310 nm, the value of $$Z$$ is ______.

[Given: $$hc = 1240$$ eV-nm and $$Rhc = 13.6$$ eV, where $$R$$ is the Rydberg constant, $$h$$ is the Planck's constant and $$c$$ is the speed of light in vacuum]

The photon that strikes the metal comes from the electronic transition $$n = 4 \rightarrow n = 3$$ in a hydrogen-like ion of atomic number $$Z$$.

Step 1 : Energy of the photon
For a hydrogen-like ion, the energy of level $$n$$ is $$E_n = -\,\dfrac{Z^{2} \, 13.6\ \text{eV}}{n^{2}}$$. Hence for the transition $$4 \rightarrow 3$$

$$\Delta E \;=\; Z^{2}\,13.6\left(\dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}\right) = Z^{2}\,13.6\Bigl(\dfrac{1}{9}-\dfrac{1}{16}\Bigr) = Z^{2}\,13.6\Bigl(\dfrac{16-9}{144}\Bigr) = Z^{2}\,13.6 \times \dfrac{7}{144}$$

Compute the numerical factor:

$$13.6 \times \dfrac{7}{144}= \dfrac{95.2}{144}=0.661111\;\text{eV}$$

Therefore the photon energy is

$$E_{\text{photon}} = 0.661111\,Z^{2}\ \text{eV} \qquad -(1)$$

Step 2 : Work function of the target metal
Threshold wavelength $$\lambda_0 = 310\ \text{nm}$$ gives the work function

$$\Phi = \dfrac{hc}{\lambda_0} = \dfrac{1240\ \text{eV·nm}}{310\ \text{nm}} = 4.00\ \text{eV}$$

Step 3 : Use photoelectric equation
Maximum kinetic energy of emitted electrons is $$K_{\max}=1.95\ \text{eV}$$. Hence

$$E_{\text{photon}} = \Phi + K_{\max} = 4.00 + 1.95 = 5.95\ \text{eV} \qquad -(2)$$

Step 4 : Equate $$E_{\text{photon}}$$ from (1) and (2)

$$0.661111\,Z^{2} = 5.95$$

$$Z^{2} = \dfrac{5.95}{0.661111} \approx 9.00$$

$$\Rightarrow\; Z \approx 3$$

Thus the atomic number of the hydrogen-like ion is 3.

Final Answer: $$Z = 3$$