Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
(i) Estimate number of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?

(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was observed instantaneously?

0 5 Views | Posted 3 months ago
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    Answered by

    Pallavi Pathak | Contributor-Level 10

    3 months ago

    Explanation-number of photon emitted per second n= phc?=p?hc=20*5000*10-106.62*10-34*3*108

    =5*1019s-1

    (ii) E=hc /? = 6.62*10-34*3*1085000*10-10*1.6*10-19=2.48eV  this enegy is greater than 2 so emission is possible

    (iii) work function ? = p4?d2*?r2?t = ?o

    ?t = 4?d2pr2 = 4*2*16*1.6*10-19*2220*(1.5*10-10)-2=28.4s

    (iv) N= n?r24?d2*?t

     = 5*1019*(1.5*10-10)2*28.44*(2)2 =2

    (v) as the time of emission is 11.04s so photoelectric is not spontaneous.

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