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`        light of wavelength 5000 *10-10  falls on metal surface of work function 1.9 eV. find (1)the energy of photons in eV (2)the K.E of photoelectrons (3)the stopping potential. `
2 years ago

Arun
23031 Points
```							Dear Kashish Here we will use the relation  E = KE + hν0  (1)where E is the energy of the incident photon, KE is the energy of the emitted photon and hν0is the work function or the minimum energy required an electron from the surface of the metal. Now as E = hν = (hc) / λwhere h is the planck's constant, c is the speed of light, λ is its wavelength of radiation and ν is its frequency. Thus the energy of the incident photon is  E = (6.63 X 10-34 X 3 X 10-8) / (5000 X 10-10) = 3.978 X 10-19 JNow 1 eV = 1.6 X 10-19 JSo 1 J = 1 / (1.6 X 10-19) eVTherefore 3.978 X 10-19 J = (1 X 3.978 X 10-19) / (1.6 X 10-19) = 2.486 eVHence the energy of the photon is 2.486 eV. Now we are given that the work function is 1.9 eV. Substituting the above value and the value of work function in equation (1), we get KE = E - hν0   = 2.486 - 1.9 = 0.586 eVSo the kinetic energy of the emitted photon is 0.586 eV.  RegardsArun (askIITians forum expert)
```
2 years ago
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