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The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellowgreen colour and about 760 nm for red colour.

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Anonym0us
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(a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum? (Take h = 6.63×10–34 J s and 1 eV = 1.6×10 –19J.)


(b) From which of the photosensitive materials with work functions listed in Table 11.1 and using the results of (i), (ii) and (iii) of (a), can you build a photoelectric device that operates with visible light?

Qalaxia Master Bot
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I found an answer from simple.wikipedia.org

Photoelectric effect - Simple English Wikipedia, the free encyclopedia


The photoelectric effect has helped physicists understand the quantum nature of light ... Not every electromagnetic wave will cause the photoelectric effect, only radiation of a ... The work function is a property of the metal and is not affected by the ... the cutoff frequency, then the emitted electron will have some kinetic energy.


For more information, see Photoelectric effect - Simple English Wikipedia, the free encyclopedia

Krishna
0

Formula of energy if photon E = hf   

          Where, h - Planck constant = 6.63* 10^{-34} J and f - frequency of waves

                         E = \frac{hc}{\lambda}                               \because f = \frac{\text{ speed of light (c)}}{\text{ Wavelength } \lambda} .

                      

Given that

  Violet colour wavelength \lambda _v = 390 nm = 390 * 10^{-9} m

  Yellow-green  colour wavelength \lambda _{yg} = 550 * 10^{-9} m

  Red colour wavelength \lambda _{r} = 760 nm = 760 * 10^{-9} m


(a)

(i) Incident photon energy for violet light

                 E = \frac{hc}{\lambda}   

                 E = \frac{6.63 * 10^{-34} * 3*10^8}{\lambda}   

                 E = \frac{1.989 * 10^{-25} Jm}{\lambda}


      Wavelength, \lambda _v = 390 nm

                         E = \frac{1.989 * 10^{-25} Jm}{390 * 10^{-9} m}

                         E = 5.10 * 10^{-19} J

                        Converting joules to electron volts

                         E = \frac{ 5.10 * 10^{-19}}{1.6*10^{-19}} J

                         E = 3.19 eV


(ii) Incident photon energy for yellow-green light

             E = \frac{1.989 * 10^{-25} Jm}{550 * 10^{-9} m}

             E = 3.62 * 10^{-19} J

             E = 2.26 eV


(iii)  Incident photon energy for red light

             E = \frac{1.989 * 10^{-25} Jm}{ 760 * 10^{-9} m}

             E = 2.62 * 10^{-19} J

             E = 1.64 eV


(b)

For a photoelectric device to operate, we need incident light energy E to be equal to or greater than the work function (\phi) ​of the material.

          

            

         E = 3.19 eV is greater than work function of photosensitive materials Na \phi = 2.75 , K \phi = 2.74 and Cs \phi = 2.14   

         Hence, the photoelectric device will operate with violet light photosensitive materials Na, K and Cs


         E = 2.26 eV > work function of photosensitive materials Cs ( \phi = 2.14 )  

         Hence, the photoelectric device will operate with yellow-green light photosensitive material Cs


         E = 1.64 eV < work function of photosensitive materials

         Thus,  the photoelectric device will not operate with red light.