A) 0.8 eV
B) 1.6 eV
C) 1.8 eV
D) 1.1 eV
Correct Answer: D
Solution :
\[r=\frac{mv}{eB}=\frac{\sqrt{2mEp}}{eB}\] [E = electron?s K.E.] [m = electron mass, e = electron charge] \[E={{E}_{p}}-\phi \]\[{{E}_{p}}=\]photon energy = 1.9 eV. \[\Rightarrow \]\[{{(reB)}^{2}}=2m[{{E}_{p}}-\phi ]\Rightarrow \phi ={{E}_{p}}-\frac{{{(reB)}^{2}}}{2m}\] \[\Rightarrow \]\[\phi =1.9eV-\frac{{{10}^{-4}}\times 1.6\times {{10}^{-19}}\times 9\times {{10}^{-8}}}{2\times 9.1\times {{10}^{-31}}}\] \[\simeq 1.9eV-0.79V=1.1eV\]
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