A) \[1:2\]
B) \[1:1\]
C) \[1:4\]
D) \[1:5\]
Correct Answer: A
Solution :
Energy of first photon \[{{E}_{1}}=1eV\] Energy of second photon \[{{E}_{2}}=2.5eV\] Work function of metal \[{{W}_{0}}=0.5\text{ }eV\] The kinetic energy available for the emission of electron is given by, \[\frac{1}{2}m{{\upsilon }^{2}}=\] energy of photon\[-\]work function \[\frac{1}{2}m{{\upsilon }^{2}}=E-{{W}_{0}}\] Or \[\upsilon =\sqrt{\frac{2E-{{W}_{0}}}{m}}\propto \sqrt{E-{{W}_{0}}}\] Hence, \[\frac{{{\upsilon }_{1}}}{{{\upsilon }_{2}}}=\sqrt{\frac{{{E}_{1}}-{{W}_{0}}}{{{E}_{2}}-{{W}_{0}}}}=\sqrt{\frac{1-0.5}{2.5-0.5}}\] \[=\sqrt{\frac{0.5}{2}}=\frac{1}{\sqrt{4}}=\frac{1}{2}\] or \[{{\upsilon }_{1}}:{{\upsilon }_{2}}=1:2\]
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