JEE Advanced 2023 — Paper 1 — PHY-III solution

JEE Advanced2023Paper 1PHY-III

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: $h c = 1240$ eV-nm and $R h c = 13.6$ eV, where $R$ is the Rydberg constant, $h$ is the Planck's constant and $c$ is the speed of light in vacuum]

Your answer

Solution

Photon energy for the $n = 4 \to n = 3$ transition. For a hydrogen-like atom, $E_{n} = - \frac{13.6 Z ^{2}}{n ^{2}}$ eV. The emitted photon has energy $E_{γ} = 13.6 Z^{2} (\frac{1}{9} - \frac{1}{16}) = 13.6 Z^{2} \cdot \frac{7}{144} \approx 0.6611 Z^{2} eV .$

Work function of the target metal. From the threshold wavelength $λ_{0} = 310$ nm: $W = \frac{h c}{λ _{0}} = \frac{1240}{310} = 4.0$ eV.

Einstein's photoelectric equation. $K_{ma x} = E_{γ} - W$ , so $0.6611 Z^{2} = 1.95 + 4.0 = 5.95 \Rightarrow Z^{2} \approx 9 \Rightarrow Z = 3.$

Answer: $Z = 3$ .

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