JEE Advanced levelNumerical answerDifficulty 3 of 3as098
Use h=6.626×10−34Js, c=3.0×108ms−1, 1eV=1.6×10−19J and E1=−13.6eV. Radiation from the n=4→n=2 transition of hydrogen is incident on a metal whose threshold wavelength is 600nm. Find the maximum kinetic energy of the ejected electrons in eV (two decimals).
Numerical answer type, answer in eV, tolerance 0.02.
Answer: 0.48 eV
Photon energy =13.6(1/4−1/16)=2.55eV. Work function ϕ=hc/λ0=3.313e−19J=2.071eV. KEmax=2.55−2.071=0.48eV.