A) 2 : 1
B) 1 : 2
C) 1 : 3
D) 1 : 4
Correct Answer: B
Solution :
The maximum kinetic energy of the emitted electron \[K=hv-{{W}_{0}}\] where hv = photon energy \[{{W}_{0}}=\] work function Suppose the mass of electron is m, then in the case I \[\frac{1}{2}mv_{1}^{2}=1\,eV-0.5=0.5\,eV\] ... (i) In case II \[\frac{1}{2}mv_{2}^{2}=2.5\,eV\,-0.5\,eV=2.0\,eV\] \[\therefore \] \[\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{0.5}{2.0}=\frac{1}{4}\] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1}{2}\]You need to login to perform this action.
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