A) 1.1
B) 2.0
C) 2.2
D) 3.1
Correct Answer: A
Solution :
[a] Energy of incident light \[E\,(eV)=\frac{12375}{4000}=3.09\,eV\] |
Stopping potential is 2V so \[{{K}_{\max }}=2\,eV\] |
Hence by using \[E={{W}_{0}}+{{K}_{\max }}\]; W0\[=1.09\,eV\]\[\approx \,1.1\,eV\] |
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