Atomic Number from Electronic Transition and Photoelectric Effect

To find the value of the atomic number $Z$, we analyze the photoelectric effect and the electronic transition in the hydrogen-like atom step-by-step:

Step 1: Determine the work function of the target metal The work function $\phi$ of the target metal is related to its threshold wavelength $\lambda_0$ by: $\phi = \frac{hc}{\lambda_0}$

Given that $hc = 1240\text{ eV-nm}$ and $\lambda_0 = 310\text{ nm}$: $\phi = \frac{1240\text{ eV-nm}}{310\text{ nm}} = 4.00\text{ eV}$

Step 2: Calculate the energy of the incident photon According to Einstein's photoelectric equation: $K_{\max} = E - \phi$

where:

• $K_{\max} = 1.95\text{ eV}$ is the maximum kinetic energy of the emitted photoelectrons.
• $E$ is the energy of the incident photon.

Thus, the energy of the photon is: $E = K_{\max} + \phi = 1.95\text{ eV} + 4.00\text{ eV} = 5.95\text{ eV}$

Step 3: Relate the photon energy to the electronic transition The energy $E$ of a photon emitted during an electronic transition from level $n_2 = 4$ to level $n_1 = 3$ in a hydrogen-like atom of atomic number $Z$ is given by the Bohr formula: $E = Rhc \cdot Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

Substitute the given values $Rhc = 13.6\text{ eV}$, $n_1 = 3$, and $n_2 = 4$: $E = 13.6 \cdot Z^2 \left(\frac{1}{3^2} - \frac{1}{4^2}\right)$ $E = 13.6 \cdot Z^2 \left(\frac{1}{9} - \frac{1}{16}\right) = 13.6 \cdot Z^2 \left(\frac{16 - 9}{144}\right) = 13.6 \cdot Z^2 \cdot \frac{7}{144}$ $E = \frac{95.2}{144} Z^2\text{ eV}$

Step 4: Solve for $Z$ Equating the two expressions for the photon energy $E$: $5.95 = \frac{95.2}{144} Z^2$

$Z^2 = \frac{5.95 \times 144}{95.2}$

Since $\frac{95.2}{5.95} = 16$: $Z^2 = \frac{144}{16} = 9$

Taking the positive square root: $Z = 3$