A) \[\Delta E\] is equal to\[\Delta E{{\left( \frac{1}{4} \right)}^{th}}\]of initial energy
B) \[\Delta E\] is equal to \[{{\left( \frac{1}{2} \right)}^{th}}\]of initial energy
C) \[\Delta E\] is equal to twice of initial energy
D) \[\Delta E\] is equal to initial energy
Correct Answer: D
Solution :
\[{{E}_{1}}=\frac{hc}{{{\lambda }_{1}}}\]and\[{{E}_{2}}=\frac{hc}{{{\lambda }_{2}}}\] Clearly,\[{{\lambda }_{1}}=2{{\lambda }_{2}}\Rightarrow {{\lambda }_{2}}=\frac{{{\lambda }_{1}}}{2}\] \[\therefore \] \[{{E}_{2}}-{{E}_{1}}=\frac{hc}{{{\lambda }_{2}}}-\frac{hc}{{{\lambda }_{1}}}\] \[\Rightarrow \] \[\Delta E=\frac{2hc}{{{\lambda }_{1}}}-\frac{hc}{{{\lambda }_{1}}}=\frac{hc}{{{\lambda }_{1}}}\] \[\Rightarrow \] \[\Delta E={{E}_{1}}\]You need to login to perform this action.
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