A) 1:2
B) 1:1
C) 1: 5
D) 1:4
Correct Answer: A
Solution :
\[{{\operatorname{E}}_{1}}=leV~~~~~{{\phi }_{1}}={{\phi }_{2}} = 0.5eV\] \[{{\operatorname{E}}_{2}}=2.5eV\] Equation of photoelectric effect \[{{\operatorname{E}}_{1}}={{\phi }_{1}}+ K{{E}_{1}}~~, {{E}_{2}}=~\phi + KE2\] \[1=0.5+\frac{1}{2}m{{v}^{2}}_{1},2.5=0.5+\frac{1}{2}m{{v}^{2}}_{2}\] \[\frac{0.5\times 2}{\operatorname{m}}={{\operatorname{v}}^{2}}_{1},\frac{2\times 2}{\operatorname{m}}={{\operatorname{v}}^{2}}_{2}\] \[{{\left[ \frac{{{v}_{1}}}{{{v}_{2}}} \right]}^{2}}=\frac{1}{4}\] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1}{2}\]1`
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