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Chapter 11.pmd


This unit of energy is commonly used in atomic and nuclear physics. The work function ... 11.5 PHOTOELECTRIC EFFECT AND WAVE THEORY. OF LIGHT ... photon. Equation (11.2) is known as Einstein's photoelectric equation. We now see ... The dual (wave-particle) nature of light (electromagnetic radiation, in general) ...


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Apr 10, 2017 ... We are hoping the ELS program will find your interest and that you will ... Wilhelms-Universität Münster, Germany in partnership with NASA's ... METAL-‐ SILICATE PARTITIONING OF K AS A FUNCTION ... rays and solar wind for future human missions. ... OHB System as part of the Lunar Volatiles Mobile.


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Pravalika
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Given that

Wavelength of Ultraviolet light \lambda = 2271 Angstrom = 2271 * 10^{-10} m

Power of Ultraviolet light = 100 W

Stopping potential V = - 1.3 volts


Wavelength of red light \lambda _r = 6328 \text{ Angstrom } = 6328 * 10^{-10} m

Intensity = 10^5 W m^{-2}


Step 1: Get an expression for the work function

             We know that

                By conservation by law, Energy E = eV...........................(1)

                     Where, e - electric charge and V - potential of electrons  

                 Einstein's photoelectric equation

                            Energy of proton  E=h\upsilon\ -\ \phi

                                                         eV = h \frac{c}{\lambda} - \phi                              \because \text{ frequency } \upsilon = \frac{c}{\lambda}   

                                                         \phi = h \frac{c}{\lambda} - eV      


Step 2: Evaluating the work function of the metal

                Planck's constant, h = 6.63* 10^{-34} J

                Electron charge, e = 1.6 * 10^{-19} C

                     \phi = 6.63* 10^{-34} * \frac{3* 10^{8}}{2271 * 10^{-10}} - (1.6 * 10^{-19}) * (1.3)

                     \phi = 8.72 *10^{-19} - 2.08 *10^{-19}

                     \phi = 6.64 * 10^{-19}   Joules

                     \phi = 4.15 eV                 \because 1 eV = 1.6*10^{-19}

                  

                Therefore, work function of the metal   \phi = 4.15 eV


Step 3: Calculating the threshold frequency of the metal  

              Relation between the work function and threshold frequency

                       \phi = h \upsilon

          

              Threshold frequency \upsilon = \frac{\phi}{h} \phi = h \upsilon

                       \upsilon = \frac{6.64 * 10^{-19}}{6.63* 10^{-34}}   

                       \upsilon = 1.0006 * 10^{15} Hz

                      

            Hence, threshold frequency \upsilon = 1.0006 * 10^{15} Hz

  

Step 4: Frequency of the red light estimation

                   \lambda _r = 6328 \text{ Angstrom } = 6328 * 10^{-10} m

              

                Frequency of the red light \upsilon_r = \frac{c}{\lambda_r}   

                         \upsilon_r = \frac{ 3* 10^8}{6328 * 10^{-10}}

                         \upsilon_r = 4.74 * 10^{14} Hz


                Thus, frequency of the red light \upsilon_r = 4.74 * 10^{14} Hz

                            

                Therefore, The photocell will not react to this red light, however strong its intensity may be. since frequency of the red

                light is less than threshold frequency (   \upsilon > \upsilon_r )