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← JEE Advanced 2023 · Paper 1
PhysicsInteger+4Question 25 of 51

JEE Advanced 2023 · Paper 1, Physics question 8

A Hydrogen-like atom has atomic number ZZ. Photons emitted in the electronic transitions from level n=4n = 4 to level n=3n = 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 ZZ is _____.

[Given: hc=1240hc = 1240 eV-nm and Rhc=13.6Rhc = 13.6 eV, where RR is the Rydberg constant, hh is the Planck's constant and cc is the speed of light in vacuum]

Answer and worked solution

Official answer: 3

Photon energy for the 4→34\to3 transition:
E=13.6 Z2(19−116)=13.6 Z2⋅7144 eV.E=13.6\,Z^2\left(\frac19-\frac1{16}\right)=13.6\,Z^2\cdot\frac7{144}\ \text{eV}.

Work function. ϕ=1240310=4\phi=\dfrac{1240}{310}=4 eV, so
E=Kmax⁡+ϕ=1.95+4=5.95 eV.E=K_{\max}+\phi=1.95+4=5.95\ \text{eV}.

Solve.
Z2=5.95×14413.6×7=9  ⟹  Z=3.Z^2=\frac{5.95\times144}{13.6\times7}=9\implies Z=3.