Modern Physics 2 Question 7

7. A metal plate of area 1×104m2 is illuminated by a radiation of intensity 16mW/m2. The work function of the metal is 5eV. The energy of the incident photons is 10eV and only 10 of it produces photoelectrons. The number of emitted photoelectrons per second and their maximum energy, respectively will be (Take, 1eV=1.6×1019J)

(a) 1011 and 5eV

(b) 1012 and 5eV

(c) 1010 and 5eV

(d) 1014 and 10eV

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Solution:

  1. We know that, intensity of a radiation I with energy ’ E ’ incident on a plate per second per unit area is given as

I=dEdA×dtdEdt=IdA or IA

i.e., energy incident per unit time =IA

Substituting the given values, we get

dEdt=16×103×1×104dEdt=16×107W

Using Einstein’s photoelectric equation, we can find kinetic energy of the incident radiation as

E=12mv2+φ or ( Here, φ is work =KE+φKE=Eφ=10eV5eVKE=5eV

Now, energy per unit time for incident photons will be

E=NhνdEdt=hνdNdt or hνN˙

From Eqs. (i) and (iii), we get

hN˙=16×107 or EN˙=16×107 But E=10eV, so N˙(10×1.6×1019)=16×107N˙=1012

Only 10 of incident photons emit electrons.

So, emitted electrons per second are

10100×1012=1011



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