A) 3600\[\mu =0.3\]
B) 2400 \[\therefore \]
C) 1800\[F=0.3\times 250=75N.\]
D) 1200 \[\overrightarrow{\mathbf{F}}\]
Correct Answer: C
Solution :
The minimum energy required for the emission of photoelectron from a metal is called work function of that metal, \[W=\frac{hc}{\lambda }\] where h is Plancks constant, c is speed of light, \[\lambda \] is wavelength. \[\Rightarrow \] \[\lambda =\frac{hc}{W}\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6.825\times 1.6\times {{10}^{-19}}}\] \[=1.8\times {{10}^{-7}}\] \[=1800\times {{10}^{-10}}m\] \[=1800\overset{\text{o}}{\mathop{\text{A}}}\,\]
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