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]

 To determine the atomic number 

$Z$ of the hydrogen-like atom, we'll follow these steps:

1. Calculate the energy of the emitted photon:

   The energy difference between two levels $n_i$ and $n_f$ in a hydrogen-like atom is given by:

   $$E=Z^2\cdot R_{h}c(\frac{1}{n_f^2}-\frac{1}{n_i^2})$$

   where $R_{h}c=13.6\text{eV}$.

   For the transition from $n=4$ to $n=3$:

   $$E=Z^2\cdot 13.6\text{eV}(\frac{1}{3^2}-\frac{1}{4^2})=Z^2\cdot 13.6\text{eV}(\frac{1}{9}-\frac{1}{16})$$$$E=Z^2\cdot 13.6\text{eV}\left(\frac{16-9}{144}\right)=Z^2\cdot 13.6\text{eV}\cdot \frac{7}{144}$$$$E=Z^2\cdot 0.661\text{eV}$$

2. Relate the photon energy to the photoelectric effect:

   The maximum kinetic energy $K_{\text{max}}$ of the photoelectrons is given by:

   $$K_{\text{max}}=E-\varphi$$

   where $\varphi$ is the work function of the metal.

   The work function can be calculated from the threshold wavelength ${\lambda }_0=310\text{nm}$:

   $$\varphi =\frac{hc}{{\lambda }_0}=\frac{1240\text{eV-nm}}{310\text{nm}}=4\text{eV}$$

   Given $K_{\text{max}}=1.95\text{eV}$:

   $$1.95\text{eV}=E-4\text{eV}$$$$E=1.95\text{eV}+4\text{eV}=5.95\text{eV}$$

3. Solve for $Z$:

   From the energy of the photon:

   $$5.95\text{eV}=Z^2\cdot 0.661\text{eV}$$$$Z^2=\frac{5.95}{0.661}\approx 9$$$$Z=\sqrt{9}=3$$

Final Answer:

$$\boxed{{\displaystyle 3}}$$