A) 532 MW
B) 352 MW
C) 235MW
D) 325 MW
Correct Answer: C
Solution :
[c] Total number of ions in the rod \[N=\left( 2\times {{10}^{25}} \right)\times \left[ \frac{\pi }{4}\times {{\left( {{10}^{-2}} \right)}^{2}} \right]\times \left( 5\times {{10}^{-2}} \right)\] \[=7.85\times {{10}^{19}}\] The energy of excitation \[E=N\times \frac{hc}{\lambda }\] \[\left( 7.85\times {{10}^{19}} \right)\times \frac{\left( 6.6\times {{10}^{-34}} \right)\times \left( 3\times {{10}^{8}} \right)}{6.6\times {{10}^{-7}}}\] \[=23.55J\] Average power P \[=\frac{E}{t}=\frac{23.55}{{{10}^{-7}}}=23.55\times {{10}^{7}}W=235MW\]You need to login to perform this action.
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