A) \[\approx 0.3\times {{10}^{6}}\,m{{s}^{-1}}\]
B) \[\approx 6\times {{10}^{5}}\,m{{s}^{-1}}\]
C) \[\approx 0.6\times {{10}^{6}}\,m{{s}^{-1}}\]
D) \[\approx 61\times {{10}^{3}}\,m{{s}^{-1}}\]
Correct Answer: B , C
Solution :
\[{{\lambda }_{0}}=3250\times {{10}^{-10}}\,m\] \[\lambda =2536\times {{10}^{-10}}\,m\] \[\text{o }\!\!|\!\!\text{ =}\frac{1242\,eV-nm}{325\,nm}=3.82\,eV\] \[hv=\frac{1242\,eV-nm}{253.6\,nm}=4.89\,eV\] \[K{{E}_{\max }}=(4.89-3.82)\,eV=1.077\,eV\] \[\frac{1}{2}m{{v}^{2}}=1.077\times 1.6\times {{10}^{-19}}\] \[v=\sqrt{\frac{2\times 1.077\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}}\] \[v=0.6\times {{10}^{6}}\,m/s\]Solution :
\[{{\lambda }_{0}}=3250\times {{10}^{-10}}\,m\] \[\lambda =2536\times {{10}^{-10}}\,m\] \[\text{o }\!\!|\!\!\text{ =}\frac{1242\,eV-nm}{325\,nm}=3.82\,eV\] \[hv=\frac{1242\,eV-nm}{253.6\,nm}=4.89\,eV\] \[K{{E}_{\max }}=(4.89-3.82)\,eV=1.077\,eV\] \[\frac{1}{2}m{{v}^{2}}=1.077\times 1.6\times {{10}^{-19}}\] \[v=\sqrt{\frac{2\times 1.077\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}}\] \[v=0.6\times {{10}^{6}}\,m/s\]
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