Question

The number of photons emitted by a $60W$ bulb per second, if $10$% of the electrical energy supplied to an incandescent light bulb is radiated as visible light (\lambda = 6000 A), is:

• A. $1.8\times {10}^{19}$
• B. $1.8\times {10}^{16}$
• C. $1.8\times {10}^{11}$
• D. $1.8\times {10}^{21}$

Answer 1

The correct option is A $1.8\times {10}^{19}$

electrical energy supplied in one sec = 60 J

(as power = 60 W)

10% of this energy $=60\times \frac{10}{100}=6J$

Only 6J of energy is converted to light energy

energy of each proton $=\frac{hc}{\lambda }=\frac{hc}{A}$

$=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{6000\times {10}^{-10}}$

$=3.313\times {10}^{-19}$ J

No. of photons $=\frac{energy/persec}{energyofphoto}$

$=\frac{6}{3.313\times {10}^{-19}}$

$=1.8\times {10}^{-19}$