JEE Advanced2015PhysicsDual Nature of MatterActual
Consider a hydrogen atom with its electron in the n t h orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is h c = 1242 e V n m
Correct answer
0
Step-by-step solution
E P h = h c λ = 1242 90 = 13.8 e V E P h = ∆ E + ( K . E . ) 13.8 = ∆ E + 10.4 ∆ E = 3.4 e V So electron initially was in n = 2