A) \[1.2\times {{10}^{14}}Hz\]
B) \[8\times {{10}^{15}}Hz\]
C) \[1.2\times {{10}^{16}}Hz\]
D) \[4\times {{10}^{12}}Hz\]
Correct Answer: B
Solution :
Sol. \[K{{E}_{1}}=h{{v}_{1}}-h{{v}_{0}}\] \[K{{E}_{2}}=h{{v}_{2}}-h{{v}_{0}}\] \[\because \] \[K{{E}_{1}}=2-K{{E}_{2}}\] \[\therefore \] \[h{{v}_{1}}-h{{v}_{0}}=2(h{{v}_{2}}-h{{v}_{0}})\] \[h{{v}_{0}}=2h{{v}_{2}}-h{{v}_{1}}\] \[{{v}_{0}}=2{{v}_{2}}-{{v}_{1}}\] \[=2(2\times {{10}^{16}})-(3.2\times {{10}^{16}})\] \[=0.8\times {{10}^{16}}Hz\] \[=8\times {{10}^{15}}Hz\]
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