A) ultraviolet region
B) visible region
C) infrared region
D) X-ray region
Correct Answer: A
Solution :
| According to laws of photoelectric effect \[K{{E}_{\max }}=E-\phi \] where \[\phi \] is work function and \[K{{E}_{\max }}\] maximum kinetic energy of photoelectron. |
| \[\therefore \] \[hv=e{{V}_{0}}+\phi \] |
| or \[hv=5\,eV+6.2\,eV=11.2\,eV\] |
| \[\therefore \] \[\lambda =\left( \frac{12400}{11.2} \right){\AA}\approx 1000{\AA}\] |
| Hence, the radiation lies in ultraviolet region. |
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