JEE Main202623 January 2026Evening ShiftChemistryStructure of AtomActual
The work functions of two metals (M_ A . and .M_ B ) are in the 1: 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of M_ A : M_ B is in the ratio of 2.642: 1 . The work functions (in eV) of M_ A and M_ B are respectively.
Options
- A1.4,2.8
- B2.3,4.6
- C1.5,3.0
- D3.1,6.2
Correct answer
B. 2.3,4.6
Step-by-step solution
The work functions satisfy W_A : W_B = 1:2 . Let W_A = W and W_B = 2W . Using the photoelectric equation KE = h - W , we have KE_A = 6 - W and KE_B = 6 - 2W . Given the kinetic energy ratio KE_A KE_B = 2.642 , we get 6-W 6-2W = 2.642 . Solving: 6 - W = 2.642(6 - 2W) gives 6 - W = 15.852 - 5.284W So 4.284W = 9.852 and W = 2.3 eV. Thus W_A = 2.3 eV and W_B = 4.6 eV.