A) greater than\[v{{\left( \frac{4}{3} \right)}^{1/2}}\]
B) \[v{{\left( \frac{3}{4} \right)}^{1/2}}\]
C) \[v{{\left( \frac{4}{3} \right)}^{1/2}}\]
D) less than \[v{{\left( \frac{4}{3} \right)}^{1/2}}\]
Correct Answer: A
Solution :
According to photoelectric equation, \[hv-{{\omega }_{0}}=\frac{1}{2}mv_{\max }^{2}\] \[\frac{hc}{\lambda }-\frac{h}{{{\lambda }_{0}}}=\frac{1}{2}mv_{\max }^{2}\] \[hc\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda \,.\,\,{{\lambda }_{0}}} \right)=\frac{1}{2}mv_{\max }^{2}\] \[{{v}_{\max }}=\sqrt{\frac{2hc}{m}\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda \,.\,\,{{\lambda }_{0}}} \right)}\] When wavelength is \[\lambda \] and velocity is v, then \[v=\sqrt{\frac{2hc}{m}\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda \,.\,\,{{\lambda }_{0}}} \right)}\] ... (i) When wavelength is \[\frac{3\lambda }{4}\] and velocity is v, then \[v=\sqrt{\frac{2hc}{m}\left[ \frac{{{\lambda }_{0}}-(3\lambda /4)}{(3\lambda /4)\times {{\lambda }_{0}}} \right]}\] ?. (ii) Divide Eq. (ii) by Eq. (i), we get \[\frac{v}{v}=\sqrt{\frac{{{\lambda }_{0}}-3\lambda /4}{\frac{3\lambda {{\lambda }_{0}}}{4}}\times \frac{\lambda {{\lambda }_{0}}}{{{\lambda }_{0}}-\lambda }}\] \[v=v{{\left( \frac{4}{3} \right)}^{\frac{1}{2}}}\sqrt{\frac{[{{\lambda }_{0}}-(3\lambda /4)]}{\lambda {{\lambda }_{0}}}}\] \[\Rightarrow \] \[v>v{{\left( \frac{4}{3} \right)}^{\frac{1}{2}}}\]
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