A) \[<2.8\times {{10}^{-10}}m\]
B) \[<2.8\times {{10}^{-9}}m\]
C) \[\le 2.8\times {{10}^{-9}}m\]
D) \[\le 2.8\times {{10}^{-12}}m\]
Correct Answer: B
Solution :
| As, energy of photon, \[E=hv\] |
| \[E=\frac{hc}{\lambda }\] |
| \[\Rightarrow \] \[E=\frac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{500\times {{10}^{-9}}}\] |
| \[\Rightarrow \] \[E=\frac{0.0397\times {{10}^{-34}}\times {{10}^{8}}}{{{10}^{-9}}}=0.0397\times {{10}^{-21}}\text{J}\] |
| \[E=\frac{0.0397\times {{10}^{-21}}}{1.6\times {{10}^{-19}}}=0.0248\times {{10}^{2}}eV\] |
| \[=2.48\,eV\] |
| According to Einstein's photoelectric emission, we have |
| \[K{{E}_{\max }}=R-W=2.48-2.28=0.2eV\] |
| For de-Broglie wavelength of the emitted electron, |
| \[{{\lambda }_{e\,\min }}=\frac{12.27A}{\sqrt{K{{E}_{\max }}(eV)}}=\frac{12.27}{\sqrt{0.2}}\] |
| \[=27.436\overset{{}^\circ }{\mathop{\text{A}}}\,\] |
| \[=27.436\times {{10}^{-10}}m\] |
| Thus, minimum wavelength of the emitted electron. |
| \[{{\lambda }_{\min }}\,=2.7436\times {{10}^{-9}}\,m\] |
| i.e., \[\lambda \ge {{\lambda }_{\min }}\] |
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