Question

Find the de broglie wavelength of hydrogen atom at a temperature 303 K?

Answer 1

I got `"0.145 nm"`.

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Well, we assume that for hydrogen atom, the temperature is high enough that the equipartition theorem applies and the average kinetic energy is given by:

`<< kappa >> = N/2 k_BT`

where `N` is the number of degrees of freedom, `k_B = 1.38065 xx 10^(-23) "J/particle"cdot"K"` is the Boltzmann constant, and `T` is the temperature in `"K"`.

`N = 3` because atoms can only fly forward in three dimensions, and not rotate or vibrate. One hydrogen atom would then have a kinetic energy that is its own average, so:

`<< kappa >> = 3/2 cdot 1.38065 xx 10^(-23) "J/atom"cdotcancel"K" cdot 303 cancel"K"`

`= 6.275 xx 10^(-21) "J/atom"`

The kinetic energy of a single atom is also equal to

`<< kappa >>_"single atom" = K = 1/2mv^2 = p^2/(2m)`

where `p` is the forward linear momentum, and `m` is the mass of the atom.

Therefore, its momentum is:

`p = sqrt(2m<< kappa >>)`

`= sqrt(2cdot"0.0010079 kg"//cancel"mol" xx cancel"1 mol"/(6.0221413 xx 10^(23) cancel"particles") cdot 6.275 xx 10^(-21) "kg"cdot"m"^2"/s"^2cdotcancel"atom")`

`= 4.583 xx 10^(-24) "kg"cdot"m/s"`

And now, by the de Broglie relation:

`lambda = h/p`,

where `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant and `lambda` is the wavelength in `"m"` of a mass-ive particle.

Therefore:

`color(blue)(lambda) = (6.626 xx 10^(-34) cancel"kg"cdot"m"^cancel(2)"/"cancel"s")/(4.583 xx 10^(-24) cancel("kg"cdot"m/s"))`

`= 1.446 xx 10^(-10) "m"`

`=` `color(blue)("0.145 nm")`