question_answer

Draw a graph showing the variation of stopping potential with the frequency of incident   radiation in   regarting to photoelectric effect.

(i) What does the slope of this graph represent?

(ii) How can the value of work function of the material be determined from this graph?

Or

The work function for the following metals is given, Na: 2.75 eV; K : 2.30 eV; Mo : 4.17 eV Ni : 5.15 eV Which of these metals will not give photoelectric emission for a radiation of wavelength \[3300\text{ }\overset{{}^\circ }{\mathop{A}}\,\] from a He-Cd laser placed 1m away from the photocell? What happens, if the laser is brought nearer and placed 50 cm away?

Answer:

(i) According to Einstein?s photoelectron equation, the maximum KE of photoelectron is given by \[{{K}_{\max }}=hv-{{W}_{0}}\] If \[{{V}_{0}}\] is the stopping potential, then             \[{{K}_{\max }}=e{{V}_{0}}\] \[\therefore \]      \[{{V}_{0}}=\left( \frac{h}{e} \right)v-\frac{{{W}_{0}}}{e}\] We compare this equation with the straight line equation, \[y=mx+C\] Clearly, slope of \[{{V}_{0}}\text{-v}\] graph \[=\frac{h}{e}\] (ii) The intercept on the vertical axis \[=-\frac{{{W}_{0}}}{e}\] \[\therefore {{W}_{0}}=e\times \]magnitude of the intercept on vertical axis In this way, the work function \[{{W}_{0}}\] can be determined. Or Wavelength of incident radiation,             \[\lambda =3300\overset{\text{o}}{\mathop{\text{A}}}\,=3300\times {{10}^{-12}}\,\,m\] \[\therefore \]Energy of an incident photon, \[E=\frac{hc}{\lambda }=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{3300\times {{10}^{-10}}}\] \[=\frac{6.63\times 3\times {{10}^{-18}}}{33\times 1.6\times {{10}^{-19}}}eV=3.75\,\,eV\]E As the energy of incident photon is less than the work function of Mo and Ni, so the metal Mo and Ni will not give photoelectric emission. If the lesser is brought closer, intensity of radiation increases. This does not affect the result regarding Mo and Ni, but the photoelectric current will increase for Na and K with the increase in intensity.