A) \[=0.6\times {{10}^{6}}m{{s}^{-1}}\]
B) \[=61\times {{10}^{3}}m{{s}^{-1}}\]
C) \[=0.3\times {{10}^{6}}m{{s}^{-1}}\]
D) \[=6\times {{10}^{6}}m{{s}^{-1}}\]
Correct Answer: A
Solution :
[a] Given, \[{{\lambda }_{0}}=3250\times {{10}^{-10}}m\] \[\lambda =2536\times {{10}^{-10}}m\] \[\phi =\frac{hc}{{{\lambda }_{0}}}=\frac{4.14\times {{10}^{-15}}\times 3\times {{10}^{8}}}{3250\times {{10}^{-10}}}=3.82eV\] \[hv=\frac{hc}{\lambda }=\frac{4.14\times {{10}^{-15}}\times 3\times {{10}^{8}}}{2536\times {{10}^{-10}}}=4.89eV\] According to Einstein's photoelectric equation, \[{{K}_{\max }}=hv-\phi \] \[K{{E}_{\max }}=\left( 4.89-3.82 \right)eV=1.077eV\]\[\frac{1}{2}m{{v}^{2}}=1.077\times 1.6\times {{10}^{-19}}\]\[\Rightarrow v=\sqrt{\frac{2\times 1.077\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
