Question

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

Answer 1

$\huge \mathfrak{solution}$

Given:

• Power of the bulb = 100 watt

• wavelength = 100nm = 100×10^-9m

To find :

• Number of photons emitted per second=?

As we know that :

$\boxed{E = \frac{hc}{ \lambda} }$

where ,

•h (Planck's constant)= 6.62×10^-34Js

•c=3×10^8 m/s

$E = \frac{6.62 \times {10}^{ - 34} \times 3 \times {10}^{8} }{1 \times {10}^{ - 7} } \\ \\ E = 19.86 \times {10}^{ - 34 + 15} \\ \\ E = 19.86 \times {10}^{ - 19} J \:$

Number of photons emitted per second is

$\implies \: N=P/E \\ \\ \implies \: N= \: \frac{100}{19.86 \times {10}^{ - 19} } \\ \\ \implies \: N= 5.03 \times {10}^{19}$

Number of photons = 5.03×10^19 s^-1

Answer 2

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.

$\hrulefill\hrulefill$

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

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

Putting the given values, we get,

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

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

∴ Number of photons emitted,

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

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