A) \[h=\left| \frac{\left( K.{{E}_{2}}-K.{{E}_{1}} \right)\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)}{C\left( {{\lambda }_{1}}-{{\lambda }_{2}} \right)} \right|\]
B) \[h=\left| \frac{\left( K.{{E}_{1}}-K.{{E}_{2}} \right)\left( {{\lambda }_{2}}-{{\lambda }_{1}} \right)}{C{{\lambda }_{1}}{{\lambda }_{2}}} \right|\]
C) \[h=\left| \frac{\left( K.{{E}_{1}}-K.{{E}_{2}} \right){{\lambda }_{1}}{{\lambda }_{2}}}{C\left( {{\lambda }_{2}}-{{\lambda }_{1}} \right)} \right|\]
D) None of These
Correct Answer: C
Solution :
[c] \[\frac{hc}{{{\lambda }_{1}}}-\frac{hc}{{{\lambda }_{0}}}=K.E{{.}_{1}}\text{ and }\frac{hc}{{{\lambda }_{2}}}-\frac{hc}{{{\lambda }_{0}}}=K.E{{.}_{2}}\] \[\Rightarrow \frac{hc}{{{\lambda }_{1}}}-\frac{hc}{{{\lambda }_{2}}}=K.E{{.}_{1}}-K.E{{.}_{2}}\] \[\Rightarrow hc\left[ \frac{{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{1}}{{\lambda }_{2}}} \right]=K.E{{.}_{1}}-K.E{{.}_{2}}\] \[\therefore h=\frac{\left( K.E{{.}_{1}}-K.E{{.}_{2}} \right){{\lambda }_{1}}{{\lambda }_{2}}}{c\left( {{\lambda }_{2}}-{{\lambda }_{1}} \right)}\]You need to login to perform this action.
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