A) \[=v{{\left( \frac{3}{5} \right)}^{1/2}}\]
B) \[>v{{\left( \frac{4}{3} \right)}^{1/2}}\]
C) \[<v{{\left( \frac{4}{3} \right)}^{1/2}}\]
D) \[=v{{\left( \frac{4}{3} \right)}^{1/2}}\]
Correct Answer: B
Solution :
\[E={{(KE)}_{\max }}+f\] \[\left[ \frac{hc}{\lambda }={{(KE)}_{\max }}+\phi \right]\] ?(1) \[\frac{4}{3}\frac{hc}{\lambda }=\left( \frac{4}{3}K{{E}_{\max }}+\frac{\phi }{3} \right)+\phi \] \[K{{E}_{\max }}\]for fastest emitted electrum\[=\frac{1}{2}mV{{'}^{2}}+\phi \] \[=\frac{1}{2}V{{'}^{2}}\frac{4}{3}\left( \frac{1}{2}m{{V}^{2}} \right)+\frac{\phi }{3}\]\[\]
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