A) \[6.82\text{ }eV\]
B) \[8.52\text{ }eV\]
C) \[12.5\text{ }eV\]
D) \[7.55\text{ }eV\]
Correct Answer: D
Solution :
\[B={{B}_{0}}[sin(3.14\times {{10}^{7}})ct+sin(6.28\,\,\times \,\,{{10}^{7}})ct]\] \[\phi \,\,=\,\,4.7eV\] \[{{K}_{\max }}\,=\,\,\frac{hc}{\lambda }\,-\,\phi \,\,=\,\,hf\,-\,\phi \] \[=\,\,\,\frac{6.6\,\times {{10}^{-34}}\,\times \,{{10}^{7}}\,c}{1.6\,\,\times \,\,{{10}^{-19}}}\,\,-\,\phi \] \[{{K}_{\max }}\,\,=\,\,\frac{6.6\,\times {{10}^{-27}}\,\times \,3\,\times \,{{10}^{8}}}{1.6\,\,\times \,\,{{10}^{-19}}}\,\,-\,4.7\,eV\] \[=\,\,\,\frac{19.6}{1.6}\,-\,4.7\,\] \[=\text{ }12.25-4.70=7.55\text{ }eV\]You need to login to perform this action.
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