A) \[V=12\times {{10}^{5}}m/s\]
B) \[V=1\times {{10}^{4}}m/s\]
C) \[V=3.8\times {{10}^{5}}m/s\]
D) \[V=1.6\times {{10}^{4}}m/s\]
Correct Answer: C
Solution :
[c] The intensity of electric field \[E=\frac{V}{d}=\frac{0.3}{1\times {{10}^{-6}}}=3\times {{10}^{5}}\,V/m\] The electric field retarded the electron. The retardation \[a=\frac{Ee}{m}\] The speed of the electron on entering p-side \[{{v}^{2}}={{u}^{2}}-2as={{u}^{2}}-2\frac{eE}{m}.d\] \[={{(5\times {{10}^{5}})}^{2}}-\frac{2\times 1.6\times {{10}^{-19}}\times 3\times {{10}^{5}}}{9.1\times {{10}^{-31}}}\times (1\times {{10}^{-6}})\] \[=3.8\times {{10}^{5}}\,m/s\]You need to login to perform this action.
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