A) 3eV
B) 1.2eV
C) 0.6 eV
D) 1.4 eV
Correct Answer: C
Solution :
Maximum kinetic energy of photoelectron \[{{(KE)}_{\max }}=E-\phi \] \[=1.8-1.2\] \[=0.6\,eV\] \[i.e,\] \[e{{V}_{0}}={{(KE)}_{\max }}\] or \[e{{V}_{0}}=0.6\,eV\] \[\therefore \] Stopping potential \[{{V}_{0}}=0.6\,eV.\]You need to login to perform this action.
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