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Question

Value of E(in eV) is
  1. 32.2
  2. 24.6
  3. 20.8
  4. 23.8

A
32.2
B
24.6
C
23.8
D
20.8
Solution
Verified by Toppr

Ionization Energy of atom of atomic number Z in nth orbit is 13.6Z2/n2. So work function for metal A WA=13.6/22=3.4eV
and, work function for metal B WB=13.6×22/22=13.6eV
Now, EWA=KEA
and, EWB=KEB
given that, KEA=2KEB
so, EWA=2E2WB
E=2WBWA=27.23.4=23.8eV

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