A) \[8000\,\overset{\circ }{\mathop{A}}\,\]
B) \[7000\,\overset{\circ }{\mathop{A}}\,\]
C) \[1472\,\overset{\circ }{\mathop{A}}\,\]
D) \[2950\,\overset{\circ }{\mathop{A}}\,\]
Correct Answer: D
Solution :
The minimum frequency (v) of light which can emit photoelectrons from a material is called the cut-off frequency of that material, and the corresponding wavelength \[\left( \lambda \right)\] is called cut off wavelength \[E=hv=\frac{hv}{\lambda }\] Where \[h\]Planck?s constant, \[c\]is speed of light. Given, \[E=4.2\,eV\] Also \[1.6\times {{10}^{-19}}\,J=1\,eV\] \[\therefore \] \[E=4.2\times 1.6\times {{10}^{-19}}\,J\] Hence, \[4.2\times 1.6\times {{10}^{-19}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{\lambda }\] \[\Rightarrow \] \[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4.2\times 1.6\times {{10}^{-19}}}\] \[=2946\times {{10}^{-10}}\,m\] \[\approx 2950\,\overset{\circ }{\mathop{\text{A}}}\,\] Note: If the frequency of incident light is below the cut off frequency, then no photoelectrons are emitted.
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