Question

a 100 watt bulb emits monochromatic light of wavelength 400 nm. calculate the number of photons emitted per second by the bulb

answer fsst ma​

Answer 1

Answer :

• Number of photons emitted = 2.012 × 10²⁰ s⁻¹.

Step-by-step explanation:

Given that,

• Power of bulb = 100 watt = 100 J s⁻¹.
• Wavelength, λ = 400 nm = 400 × 10⁻⁹ m.

Also,

• Speed of light, c = 3 × 10⁸ m s⁻¹.
• Planck's constant, h = 6.626 × 10⁻³⁴ J s.

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We know, energy of one photon is given by the formula:-

$\implies\rm{E=hv=hc/\lambda}$

Putting the given values, we get,

$\implies\rm{E=\dfrac{6.626\times{}10^{-34}\:Js\times{}3\times{}10^8\:ms^{-1}}{400\times{}10^{-9}\:m}}$

$\implies\rm{E=4.969\times{}10^{-19}\:J}$

∴ Number of photons emitted,

$\implies\rm{\dfrac{100\:Js^{-1}}{4.969\times{}10^{-19}\:J}}$

$\implies\bold{2.012\times{}10^{20}\:s^{-1}.}$

Answer 2

Answer:

2× 10²⁰s minus 1

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