The energy of a photon of sodium light (lambd | vedclass

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(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Using $hc = 12400  eV \cdot \mathring{A}$ and $\lambda = 589  nm = 5890  \mathring{A}$

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Using $hc = 12400  eV \cdot \mathring{A}$ and $\lambda = 589  nm = 5890  \mathring{A}$, we get:$E = \frac{12400}{5890} \approx 2.105  eV$.To find the ratio $E/kT$, we convert energy to Joules: $E = 2.105 	imes 1.6 	imes 10^{-19}  J$.The thermal energy $kT$ at $T = 300  K$ is $k = 1.38 	imes 10^{-23}  J/K$.$kT = 1.38 	imes 10^{-23} 	imes 300 = 4.14 	imes 10^{-21}  J$.Now, $\frac{E}{kT} = \frac{2.105 	imes 1.6 	imes 10^{-19}}{4.14 	imes 10^{-21}} = \frac{3.368 	imes 10^{-19}}{4.14 	imes 10^{-21}} \approx 81.35$.Rounding to the nearest integer, we get $81$.

The energy of a photon of sodium light $(\lambda = 589 nm)$ is equal to the band gap of a semiconductor material. Find the value of $E/kT$ at $300 K$.

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