InterviewSolution
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InterviewSolution
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Monochromatic radiation of wavelength `lambda_(1) = 3000 Å` falls on a photocell operating in saturating mode. The correspondin spectral sensitivity of photocell is `J = 4.8 xx 10^(-3) A//W`. When another monochromatic radiation of wavelength `lambda_(2) = 1650 Å` and power `P = 5 xx 10^(-3) W` is incident, it is found that maximum velocity of photoelectrons increases `n = 2` times. Assuming efficiency of photoelectron generation per incident photon to be same for both the cases, calculate (i) threshold wavelength for the cell. (ii) saturation current in second case. |
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Answer» Correct Answer - (i) `4125Å` (ii) `13.3 muA` Energy of photo with `lambda_(1) = 3000 Å = (hc)/(lambda_(1)) = 4.14 eV` Energy of photo with `lambda_(1) = 1650 Å = (hc)/(lambda_(2)) = 7.53 eV` Power of source `= 5 xx 10^(-3) W` `overset(.)(N)_(2) = (5 xx 10^(-3))/(7.53 xx 10^(-19) xx 1.6) = 4.15 xx 10^(15)` `overset(.)(N)_(1)` for `1W = (1)/(4.14 xx 1.6 xx 10^(-19)) = 1.5 xx 10^(18)` Current `= 4.8 xx 10^(-3) A` `rArr overset(.)(eta) = (4.8 xx 10^(-3))/(1.6 xx 10^(-19)) = 3 xx 10^(16)` `eta = (3 xx 10^(16))/(1.5 xx 10^(16)) = 2 xx 10^(-2) = 2%` `overset(.)(eta)_(2) = 4.15 xx 10^(16) xx 0.02 = 8.3 xx 10^(13)` Current `= overset(.)(n)e = 13.3 muA` Also `v_("max" 1650) = 2V_("max" 5000)` `rArr KE_("max" 1650) = 4KE_("max" 3000)` `rArr 4(4.14 - phi) = (7.53 - phi)` `rArr 163.56 - 7.53 = 3phi` `rArr 16.56 - 7.53 = 3phi` `rArr phi = (9.03)/(3) = 3.01 eV, lambda_(th) = (he)/(E) = 4126 Å` |
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