Question
What is the number of photons of light with a wavelength of $4000$ pm that provide $1J$ of energy?
✓
Answer
Energy $(E)$ of a photon $=h \nu$
Energy $\left(E_n\right)$ of ' $n$ ' photons $=n h v$
$\Rightarrow n=\frac{E_{n} \lambda}{hc}$
Where,
$\lambda=$ wavelength of light $=4000 pm =4000 \times 10^{-12} m$
$c=$ velocity of light in vacuum $=3 \times 10^8 m / s$
$h =$ Planck's constant $=6.626 \times 10^{-34} Js$
Substituting the values in the given expression of $n$ :
$n=\frac{(1) \times\left(4000 \times 10^{-12}\right)}{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^8\right)}=2.012 \times 10^{16}$
Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are $2.012 \times 10^{16}$.
Energy $\left(E_n\right)$ of ' $n$ ' photons $=n h v$
$\Rightarrow n=\frac{E_{n} \lambda}{hc}$
Where,
$\lambda=$ wavelength of light $=4000 pm =4000 \times 10^{-12} m$
$c=$ velocity of light in vacuum $=3 \times 10^8 m / s$
$h =$ Planck's constant $=6.626 \times 10^{-34} Js$
Substituting the values in the given expression of $n$ :
$n=\frac{(1) \times\left(4000 \times 10^{-12}\right)}{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^8\right)}=2.012 \times 10^{16}$
Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are $2.012 \times 10^{16}$.
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