JEE Advanced levelSingle correct optionDifficulty 3 of 3pt149
A hydrogen atom in its first excited state (n=2) absorbs a photon of wavelength 100 nm and is ionised. Take hc=1240eVnm and the ground-state ionisation energy of hydrogen as 13.6 eV. The kinetic energy of the ejected electron is
A12.4eV
B1.2eV
C15.8eV
D9.0eV
Answer: D
Photon energy = 1240/100 = 12.4 eV. Ionisation energy of n=2 is 13.6/4 = 3.4 eV. KE = 12.4 - 3.4 = 9.0 eV. Using 13.6 would give a negative value, and adding 3.4 gives the wrong 15.8.