A) 1 : 5
B) 1 : 4
C) 1 : 2
D) 1 : 1
Correct Answer: C
Solution :
According to Einsteins photoelectric equation, \[({{E}_{K}})=hv-{{\phi }_{0}}\] \[({{E}_{K}})=1-0.5=0.5\,eV\] Similarly, \[{{({{E}_{K}})}_{2}}=2.5-0.5=2eV\] \[\therefore \] \[\frac{{{({{E}_{K}})}_{1}}}{{{({{E}_{K}})}_{2}}}=\frac{1}{4}\] \[\Rightarrow \] \[\frac{\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv_{2}^{2}}=\frac{1}{4}\] \[\Rightarrow \] \[\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4}\] \[\Rightarrow \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1}{2}\]
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