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(B) The total energy $E$ of $n$ photons is given by $E = \frac{nhc}{\lambda}$.Given: $E = 1 \ J$,$\lambda = 4000 \ \mathring{A} = 4000 	imes 10^{-10}

Vedclass · Accepted Answer

$2 	imes 10^{18}$

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(B) The total energy $E$ of $n$ photons is given by $E = \frac{nhc}{\lambda}$.Given: $E = 1 \ J$,$\lambda = 4000 \ \mathring{A} = 4000 	imes 10^{-10} \ m = 4 	imes 10^{-7} \ m$,$h = 6.626 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$.Rearranging for $n$: $n = \frac{E \lambda}{hc}$.Substituting the values: $n = \frac{1 	imes 4 	imes 10^{-7}}{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}$.$n = \frac{4 	imes 10^{-7}}{19.878 	imes 10^{-26}} \approx 0.2012 	imes 10^{19} = 2.012 	imes 10^{18}$.Rounding to the nearest option,$n \approx 2 	imes 10^{18}$.

What will be the number of photons required to obtain $1 \ J$ of energy by radiation having a wavelength of $4000 \ \mathring{A}$?

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