A) 1.41 eV
B) 1.51 eV
C) 1.68 eV
D) 3.09 eV
Correct Answer: A
Solution :
[a] \[\lambda =400nm,hc=1240eV.nm,K.E.=1.68eV\] We know that, \[\frac{hc}{\lambda }-W=K.E.\Rightarrow W=\frac{hc}{\lambda }-K.E\] \[\Rightarrow W=\frac{1240}{400}-1.68=3.1-1.68=1.42eV\]You need to login to perform this action.
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