A) 3.5 eV
B) 13.6 eV
C) 6.8 eV
D) 1.5 eV
Correct Answer: D
Solution :
\[E={{W}_{0}}+{{K}_{\max }}\]. From the given data E is 6.78 eV (for l= 1824 \[{\AA}\]) or \[10.17\,eV\,\](for \[\lambda =1216\,\,{\AA}\]) \[\therefore \] \[{{W}_{0}}=E-{{K}_{\max }}=6.78-5.3=1.48\,eV\] or \[{{W}_{0}}=10.17-8.7=1.47\,eV.\]You need to login to perform this action.
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