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., \[\lambda \ge {{\lambda }_{\min }}\] |
You need to login to perform this action.
You will be redirected in
3 sec