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, \[hv-{{\omega }_{0}}=\frac{1}{2}mv_{\max }^{2}\] \[\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}=\frac{1}{2}mv_{\max }^{2}\] \[hc\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda .{{\lambda }_{0}}} \right)=\frac{1}{2}mv_{\max }^{2}\] \[{{v}_{\max }}=\sqrt{\frac{2hc}{m}\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda .{{\lambda }_{0}}} \right)}\] 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 \[\frac{v'}{v}=\sqrt{\frac{{{\lambda }_{0}}-3\lambda /4}{\frac{3\lambda {{\lambda }_{0}}}{4}}\times \frac{\lambda {{\lambda }_{0}}}{{{\lambda }_{0}}-\lambda }}\] \[v'=v{{\left( \frac{4}{3} \right)}^{\frac{1}{2}}}\sqrt{\frac{[{{\lambda }_{0}}-(3\lambda /4)]}{\lambda {{\lambda }_{0}}}}\Rightarrow v'>v{{\left( \frac{4}{3} \right)}^{\frac{1}{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
