NTA Abhyas JEE Main2020ChemistryStructure of AtomPractice
When photons of energy 4 . 25 e V strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, T A (expressed in eV) and de Broglie wavelength λ A . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4 . 20 eV i s T B = T A - 1 . 50 e V . If the de Broglie wavelength of these photoelectrons is λ B = 2 λ A , then which is not correct?
Options
- AThe work function of A is 2.25 eV
- BThe work function of B is 3.70 eV
- CT A = 2 . 00 e V
- DT B = 2 . 75 e V
Correct answer
D. T B = 2 . 75 e V
Step-by-step solution
λ= h mV V A = h m λ A ; V B = h m λ B V A V B = λ B λ A = 2 λ A λ A = 2 T A T B = m V A 2 m V B 2 = V A V B 2 = 2 2 = 4 Also T A - T B = 1 . 50 So, T B = 0 . 50 T A = 0 . 50 + 1 . 50 = 2 . 00 e V Also, 4 . 25 = h v 0 A + T A , So h v 0 A = 4 . 25 - 2 . 00 = 2 . 25 e V 4 . 20 = h v 0 B + T B , h v 0 B = 4 . 20 - 0 . 50 = 3 . 70 e V