Question

A hydrogen-like atom of atomic number Z transits from n=4 to n=3. The photon emitted from this transition is incident on a metal surface of threshold wavelength λth = 310nm. If the maximum kinetic energy of the emitted electron is 1.95ev, find the value of Z.

Answer 1

The correct answer is 3
$\Psi =\frac{hc}{\lambda_{th}}=\frac{1240}{310}=4eV$
$KE_{max}=h\upsilon-\Psi$
$1.95=h\upsilon-4\Rightarrow h\upsilon=5.95eV$
The energy of the photon emitted due to electron transition: $\Delta E=13.6eV\,Z^2(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}})$
$5.95\,eV=13.6\,eV(Z)^2(\frac{1}{3^2}-\frac{1}{4^2})$
Z=3