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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 11 of 67
Marks: +1, -0
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 µm. Calculate the rate of emission of quanta per second.
Solution:  
Power of bulb = 25 watt = worktime\frac{\text{work}}{\text{time}} = 25 Js−1\mathrm{Js}^{-1}
Energy of photon (E) = hcλ\frac{hc}{\lambda} = 6.626×10−34×3×1080.57×10−6\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.57 \times 10^{-6}} [Since 1 µm = 10−610^{-6} m]
Rate of emission of quanta per second = 2524.87×10−20\frac{25}{24.87 \times 10^{-20}} = 0.7169 × 102010^{20}
= 7.169 × 1019s−110^{19} \mathrm{s}^{-1}
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