A) \[v\ {{(3/4)}^{1/2}}\]
B) \[v\ {{(4/3)}^{1/2}}\]
C) Less than \[v\ {{(4/3)}^{1/2}}\]
D) Greater than \[v\ {{(4/3)}^{1/2}}\]
Correct Answer: D
Solution :
\[h\nu -{{W}_{0}}=\frac{1}{2}mv_{\max }^{2}\Rightarrow \frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}=\frac{1}{2}mv_{\max }^{2}\] \[\Rightarrow hc\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)=\frac{1}{2}mv_{\max }^{2}\]\[\Rightarrow {{v}_{\max }}=\sqrt{\frac{2hc}{m}\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)}\] 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 equation (ii) by (i), we get \[\frac{v'}{v}=\sqrt{\frac{[{{\lambda }_{0}}-(3\lambda /4)]}{\frac{3}{4}\lambda {{\lambda }_{0}}}\times \frac{\lambda {{\lambda }_{0}}}{{{\lambda }_{0}}-\lambda }}\] \[v'=v{{\left( \frac{4}{3} \right)}^{1/2}}\sqrt{\frac{[{{\lambda }_{0}}-(3\lambda /4)]}{{{\lambda }_{0}}-\lambda }}\] i.e. \[v'>v{{\left( \frac{4}{3} \right)}^{1/2}}\]
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