JEE Advanced 2023 Paper 1 · Q08 · Photoelectric Effect
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]
Reveal answer + step-by-step solution
Correct answer:3
Solution
Work function from threshold wavelength: $\phi = hc/\lambda_{\text{th}} = 1240/310 = 4$ eV. From the photoelectric equation, the incident photon energy equals $E_{\gamma} = \phi + KE_{\max} = 4 + 1.95 = 5.95$ eV. This photon comes from the $n=4 \to n=3$ transition in a hydrogen-like atom, so by the Bohr formula $$E_{\gamma} = R hc\, Z^2\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = 13.6\, Z^2 \cdot \frac{7}{144}\ \text{eV}.$$ Setting this equal to $5.95$ eV gives $Z^2 = \dfrac{5.95 \times 144}{13.6 \times 7} = 9$, hence $Z = 3$.
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