A) 1 : 5
B) 1 : 4
C) 1 : 2
D) 1 : 1
Correct Answer: C
Solution :
If E is the energy of incident photon and \[{{W}_{0}}\] the work function, then \[E-{{W}_{0}}=\] available energy \[E-{{W}_{0}}=\frac{1}{2}m{{v}^{2}}\] \[v=\sqrt{\frac{2(E-{{W}_{0}})}{m}}\] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{1-0.5}{2.5-0.5}}\] \[=\sqrt{\frac{0.5}{2}}=\frac{1}{2}\]
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