A) ultraviolet region
B) infrared region
C) visible region
D) X-ray region
Correct Answer: A
Solution :
The threshold energy,\[h{{v}_{0}}=6.2\text{ }eV\]and the stopping potential\[e{{V}_{0}}=5\text{ }eV\] From Einstein's photoelectric equation, \[hv=h{{v}_{0}}+e{{V}_{0}}=6.2+5=11.2\,eV\] \[(\because hv={{\phi }_{0}}+KE)\] \[\Rightarrow \] \[\frac{hc}{\lambda }=11.2\,eV;\]where,\[\lambda \]is the wavelength. \[\therefore \]\[\lambda =\frac{hc}{112(eV)}=1108.9{\AA}\] Which belongs to ultraviolet region.
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