Question

What is the energy of a proton with a wavelength of 700.0 nm?

Answer 1

I got `1.672 xx 10^(-9)` `"eV"`.

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Read the question carefully, and you should see it says proton... That is, a mass-ive particle with rest mass `m_p = 1.673 xx 10^(-27) "kg"`.

Objects with a rest mass follow the de Broglie relation:

`lambda = h/(mv) = h/p`

and have kinetic energy given by

`K = 1/2 mv^2 = p^2/(2m)`,

where:

• `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant.
• `p = mv` is the linear momentum, `m` is the mass, and `v` is the speed.

Thus, its kinetic energy is related to the wavelength as follows:

`p = h/lambda`

`=>` `color(green)(barul(|stackrel(" ")(" "K = (h//lambda)^2/(2m)" ")|))`

Thus, the kinetic energy is:

`K = ((6.626 xx 10^(-34) "kg"cdot"m"^2"/s")/(700.00 xx 10^(-9) "m"))^2/(2(1.673 xx 10^(-27) "kg"))`

`= 2.678 xx 10^(-28) "J"`

or

`color(blue)(K) = 2.678 xx 10^(-28) cancel"J" xx "1 eV"/(1.602 xx 10^(-19) cancel"J")`

`= color(blue)(1.672 xx 10^(-9) "eV")`

(We can only do this because protons have approx. the same charge magnitude as electrons. "Electron"-volt does not only apply to electrons.)