• # question_answer 28) The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface is 5V. The incident radiation lies in: A) X-ray region                  B) ultraviolet regionC) infrared region   D) visible region

Maximum kinetic energy of electrons ${{(KE)}_{max.}}=e{{V}_{s}}=5eV$ According to Einstein?s photoelectric equation $E={{(KE)}_{max.}}\,+\,\,\phi$ where E = Energy of incident radiation. $\phi =$ Work function $E=5+6.2=11.2\text{ }eV$ Wavelength of incident radiation, $\lambda =\frac{hc}{E}$ $=\frac{1.240\times {{10}^{-6}}}{11.2}\,eV-m$ $=\frac{1.240\times {{10}^{-6}}}{11.2}\,=0.1107\times {{10}^{-6}}\,m$ $=1107\,\,\times {{10}^{-10}}m=1107\text{ }\overset{{}^\circ }{\mathop{A}}\,$ The incident radiation lies in the ultraviolet region