A) 1.6 eV
B) 1.1 eV
C) 0.8 eV
D) 1.8 eV
Correct Answer: B
Solution :
| [b] \[\phi ={{E}_{ph}}-{{\left( KE \right)}_{\max }}\] |
| \[p=\sqrt{2mk}\] |
| \[\Rightarrow k=\frac{{{p}^{2}}}{2m}\] |
| \[r=\frac{p}{eB}\] |
| \[\Rightarrow k=\frac{{{r}^{2}}{{e}^{2}}{{B}^{2}}}{2m}\] |
| \[=\frac{12420}{6561}-\frac{{{r}^{2}}e{{B}^{2}}}{2m}\](ln eV) |
| \[=1.89\left( eV \right)-\frac{\left( {{10}^{-4}} \right)\left( 1.6\times {{10}^{-19}} \right)9\times {{10}^{5}}}{2\times 9.07\times {{10}^{-31}}}\] |
| \[=\left( 1.89-0.79 \right)eV=1.1eV\] |
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