A) \[v\,{{(3/4)}^{1/2}}\]
B) \[v\,{{(4/3)}^{1/2}}\]
C) less than \[v\,{{(4/3)}^{1/2}}\]
D) greater than \[v\sqrt{\frac{4}{3}}\]
Correct Answer: D
Solution :
| Einsteins photoelectric equation is given |
| \[{{E}_{k}}=E-W\] |
| but \[{{E}_{k}}=\frac{1}{2}m{{v}^{2}}\] |
| and \[E=\frac{hc}{\lambda }\] |
| \[\therefore \] \[\frac{1}{2}m{{v}^{2}}=\frac{hc}{\lambda }-W\] |
| Suppose v be the new speed, when \[\lambda \] is changed to \[\frac{3\lambda }{4}\], |
| \[\therefore \] \[\frac{1}{2}mv{{'}^{2}}=\frac{hc}{(3\lambda /4)}-W\] |
| or \[\frac{1}{2}mv{{'}^{2}}=\frac{4}{3}\frac{hc}{\lambda }-W\] |
| Dividing Eq. (ii) by Eq. (i), we get |
| \[\frac{v{{'}^{2}}}{{{v}^{2}}}=\frac{\frac{4}{3}\frac{hc}{\lambda }-W}{\frac{hc}{\lambda }-W}\] |
| \[=\frac{\frac{4}{3}\frac{hc}{\lambda }-\frac{4}{3}W+\frac{1}{3}W}{\frac{hc}{\lambda }-W}\] |
| \[=\frac{4}{3}+\frac{W}{3\left( \frac{hc}{\lambda }-W \right)}>\frac{4}{3}\] |
| \[\therefore \] \[\frac{v'}{v}>\sqrt{\frac{4}{3}}\] |
| or \[\sqrt{\frac{4}{3}}\,v\] |
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