The work functions of two metals (MA and MB) are in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of MA : MB is in the ratio of 2.642 : 1. The work functions (in eV) of MA and MB are respectively.
- 3.1, 6.2
- 1.5, 3.0
- 2.3, 4.6
- 1.4, 2.8
Answer
(3) 2.3, 4.6
Solution
Let A = , then B = 2 .
KE = h -
6 - 6 - 2 = 2.642
6 - = 15.852 - 5.284
4.284 = 9.852
= 2.3 eV
A = 2.3 eV and B = 4.6 eV
Answer: (3)