Question

please solve question no 1 and 5

Answer 1

(5). Energy of light absorbed in an photon = hcλabsorbedhcλabsorbed

Let n1n1 photons are absorbed

∴∴ Total energy absorbed = n1hcλabsorbedn1hcλabsorbed

Now E of light re-emitted out in one photon = hcλemittedhcλemitted

Let n2n2 photons are re-emitted then

∴∴ Total energy re-emitted out = n2hcλemittedn2hcλemitted

Eabsorbed×47100=Ere−emittedoutEabsorbed×47100=Ere−emittedout

hcλabsorbed×n1×47100=n2×hcλemittedhcλabsorbed×n1×47100=n2×hcλemitted

n2n1=47100×λemittedλabsorbedn2n1=47100×λemittedλabsorbed

=47100×50804530=47100×50804530

n2n1n2n1 = 0.527



(2). Power of bulb, P = 25 Watt = 25 Js–1

Energy of one photon, E = hν

Substituting the values in the given expression of E:


E = 34.87 × 10–20 J

Rate of emission of quanta per second