JEE MAIN - Chemistry (2021 - 1st September Evening Shift - No. 20)

A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x $$\times$$ 1020. The value of x is __________. [Given : h = 6.63 $$\times$$ 10$$-$$34 Js and c = 3.0 $$\times$$ 108 ms$$-$$1]
Answer
2

Explanation

Energy of photon is given as $$E = {{nhc} \over \lambda }$$ .... (i)

where, E = energy of photon (50 W),

n = number of photon

h = Planck's constant (6.63 $$\times$$ 10$$-$$34 Js)

c = speed of light (3 $$\times$$ 108 m/s)

$$\lambda$$ = wavelength of light (795 $$\times$$ 10$$-$$9 m)

E = 50W = 50 J = energy of photon

50 J = $${{n \times 6.63 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m/s} \over {795 \times {{10}^{ - 9}}m}}$$

$$\Rightarrow$$ $$n = {{50 \times 795 \times {{10}^{ - 9}}} \over {6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$$

$$ = 1998.49 \times {10^{17}} = 1.998 \times {10^{20}}$$

$$\Rightarrow$$ $$\approx$$ 2 $$\times$$ 1020

$$\therefore$$ x = 2

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