At what temperature would the kinetic energy of a gas molecule be equal to the energy of photon of wavelength `6000Å`? Given, Boltzmann's constant, `k = 1.38 xx 10^(-23) J .K^(-1)`, Planck's constant, `h = 6.625 xx 10^(-34) J.s`

Explanation

Let the required temperature be TK. We know, kinetic energy of a gas molecule
`= (3)/(2) kT = (3)/(2) xx 1.38 xx 10^(-23) xx T`
Again, kinetic energy of photon
`= hf = (hc)/(lamda) = (6.625 xx 10^(-34) xx 3 xx 10^(8))/(6000 xx 10^(-10))`
Hence, `(3)/(2) xx 1.38 xx 10^(-23) xx T = (6.625 xx 10^(-34) xx 3 xx 10^(8))/(6000 xx 10^(-10))`
`:. T = (2)/(3) xx (6.625 xx 10^(-34) xx 3 xx 10^(8))/(1.38 xx 10^(-23) xx 6000 xx 10^(-10)) = 1.6 xx 10^(4) K`