Question

Question #c8745

Answer 1

`lambda = "0.0123 nm"`

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Assuming you mean `1 xx 10^4` volts, that asks you to recall the definition of the electron-volt:

One electron-volt is the amount of energy needed to push an electron through a potential difference of one volt.

So, the energy involved in this process is `10^4` `"eV"`. The kinetic energy in terms of wavelength is given by

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

(or `p = sqrt(2mK)`)

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

where:

• `h = 6.626 xx 10^(-34)` `"J"cdot"s"` is Planck's constant.
• `m` is the mass of the electron in `"kg"`.
• `v` is its speed in `"m/s"`.
• `p` is its momentum.
• `lambda` is the wavelength of the electron in `"m"`.

So, first, we convert the energy from `"eV"` to `"J"`:

`10^4 cancel"eV" xx (1.602 xx 10^(-19) "J")/(cancel"1 eV") = 1.602 xx 10^(-15)` `"J"`

And now, if we derive `lambda` as a function of `K`:

`lambda = h/(sqrt(2mK))`

`= (6.626 xx 10^(-34) "kg"cdot"m"^2"/s")/(sqrt(2cdot 9.109 xx 10^(-31) "kg" cdot 1.602 xx 10^(-15) "kg"cdot"m"^2"/s"^2)`

`= 1.227 xx 10^(-11)` `"m"`

A more useful unit here would be the `"nm"`, where `"1 nm" = 10^(-9)` `"m"`:

`color(blue)(lambda = "0.0123 nm")`

This is in the X-ray region of the EM spectrum.