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