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 .

The correct answer is 3.

Step 1: Calculate the work function (threshold energy) of the target metal
The work function ($W_0$) of the metal is related to the threshold wavelength (${\lambda }_0$) by the formula:

$W_0=\frac{hc}{{\lambda }_0}$

Given that $hc\approx 1240\text{eV·nm}$ and threshold wavelength ${\lambda }_0=310\text{nm}$:

$W_0=\frac{1240\text{eV·nm}}{310\text{nm}}$

Step 2: Calculate the energy of the incident photon
According to Einstein's photoelectric equation, the maximum kinetic energy ($K_{\text{max}}$) of the photoelectrons is given by:

$K_{\text{max}}=E-W_0$

where $E$ is the energy of the incident photon. Rearranging the equation:

$E=K_{\text{max}}+W_0$

Given $K_{\text{max}}=1.95\text{eV}$ and $W_0=4.0\text{eV}$:

$E=1.95\text{eV}+4.0\text{eV}=5.95\text{eV}$

Step 3: Relate photon energy to the electronic transition in the hydrogen-like atom
The energy levels of a hydrogen-like atom with atomic number $Z$ are given by:

$E_n=-13.6\times \frac{Z^2}{n^2}\text{eV}$

The energy emitted during the transition from $n=4$ to $n=3$ is:

$\Delta E=13.6\times Z^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)\text{eV}$

$\Delta E=13.6\times Z^2\left(\frac{1}{9}-\frac{1}{16}\right)\text{eV}$

$\Delta E=13.6\times Z^2\times \frac{7}{144}\text{eV}$

Step 4: Solve for Z
Equating the emitted transition energy $\Delta E$ to the incident photon energy $E=5.95\text{eV}$:

$13.6\times Z^2\times \frac{7}{144}=5.95$

$Z^2=\frac{5.95\times 144}{13.6\times 7}$

$Z^2=\frac{856.8}{95.2}=9$

$Z=\sqrt{9}=3$

Thus, the value of $Z$ is 3.