A) 1.49eV
B) 2.2eV
C) 3.0eV
D) 5.0eV
Correct Answer: A
Solution :
[a] \[{{K}_{\max }}=hv-h{{v}_{0}}=hc\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\] Use \[hc=1.24\times {{10}^{-6}}\left( eV \right)m\] \[hc=1.24\times {{10}^{-6}}\left( \frac{{{10}^{8}}}{18}-\frac{{{10}^{8}}}{23} \right)\] \[=\frac{1.24\times 100\times \left( 23-18 \right)}{18\times 23}=1.49eV\]
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