A) 0.041 eV
B) 0.41 eV
C) 4.1 eV
D) 41 eV
Correct Answer: B
Solution :
\[{{E}_{k}}=\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}\] \[{{E}_{k}}=\frac{4.14\times {{10}^{-15}}\times 3\times {{10}^{8}}}{{{10}^{-7}}}\left( \frac{1}{5}-\frac{1}{6} \right)\] \[{{E}_{k}}=4.14\times {{10}^{-15}}\times 3\times {{10}^{15}}\left( \frac{6-5}{30} \right)\] \[{{E}_{k}}=4.14\times 3\times \frac{1}{30}\] \[{{E}_{k}}\approx 0.14eV\]
You need to login to perform this action.
You will be redirected in
3 sec
