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  • Consider a 20 W bulb emitting light of wavelength 5000 A^{circ} and shining on a metal surface kept at a distance 2m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a

Consider a 20 W bulb emitting light of wavelength 5000 A^{\circ} and shining on a metal surface kept at a distance 2m. 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 A^{\circ} .

(i) Estimate no. 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?

Answers (1)

Given that P=20W
\lambda=5000A^{\circ}=5000*10^{-10}m\\ d=2m\\ \phi =2ev\\ r=1.5*10^{10}m\\

 i) Number of photon emitted by bulb per second=

\frac{p \lambda}{hc}=5*10^{19}sec

  ii Energy of the incident photon=

\frac{hc}{ \lambda}=2.48\;eV

 iii Time required by the atomic disk to receive energy is=28.4 sec

 Let the time spent be \Delta T

E=P*A*\Delta t=P*\pi r^{2}\Delta t

 Energy transferred by the bulb in full solid angle to atoms=

4\pi d^{2}\phi

p*\pi r^{2}\Delta t=4 \pi d^{2}\phi       \Delta t=\frac{4d^{2}\phi}{Pr^{2}}=\frac{4*2*2*2*1.6*10^{-10}}{20*1.5*1.5*10^{-10}*10^{-10}}sec=\frac{12.8*10^{-19+20}}{5*2.25}

 (iv) Number of photons received by the atomic disk=N

\frac{n_{1}\pi r^{2}\Delta t}{4\pi d^{2}}=\frac{n_{1}r^{2}\Delta t}{4d^{2}}\\ \\ =\frac{5 *10^{19}*1.5*1.5*10^{-20}*11.4}{4*2*2}

\approx 0.80\cong 1 photon per atom

   (iv) We say that photoelectric emission is instantaneous as it involves a collision between the incident photon and free electron lasting for a very short span of time, say \leq 10^{-9} sec
 

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infoexpert24

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