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.
B. 2.3,4.6
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.
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