A) 0.9
B) 0.7
C) 0.5
D) 0.1
Correct Answer: C
Solution :
Energy of the radiation\[=E=\frac{hc}{\lambda }\] Inserting the values we get \[E=\frac{1242\,e\forall -nm}{2480\,nm}=0.5\,e\forall \] Hence, the correction option is [c].You need to login to perform this action.
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