QJEEM21Ch5eb2: JEE Main 2021 Chemistry - Class 11 JEEM21Ch5eb2 Solution: A 50 watt bulb emits monochromatic red light of wavelength of…

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