Question

What is the kinetic energy for one electron whose wavelength is `"650 nm"`? Its rest mass is `9.109 xx 10^(-31) "kg"`.

Answer 1

An example of De Broglie equaton.........................

Explanation:

The De Broglie equation states that the

`"KE" = (1/2)"mv"^2`

`"Where KE is kinetic energy"`
`"Where m is mass and is a constant for each particle"`
`"Where v is velocity"`

`2"KE" = "mv"^2`

`2"KEm" = "m"^2"v"^2`

`"mv" = "p"`

`therefore "2KEm" = "p"^2`

`therefore p = sqrt("2KEm")`

Another equation of De Broglie's work is

`λ = "h"/"p"`

Where `"h"` is Planck's constant

So substitute the above equation in the below equation

`λ = "h"/ sqrt("2KEm")`

`sqrt("2KEm") = "h"/λ`

As electrons mass is `9.11 xx 10^(-31) "kg"`

Convert nm to m

650nm = `650xx10^-9m`

Plug in the variables

`sqrt("2KE"xx(9.11 xx 10^(-31) "kg")) = (6.626 xx 10^(-34) "kg m"^2 s^(-2) ⋅ s)/(650xx10^-9m) =`

`sqrt("2KE"xx(9.11 xx 10^(-31) "kg")) = 1.0193846xx10^(-27)" kg m⋅ s"^(-2) ⋅ "s"`

But you can't solve this equation so

`2"KE" = ("mv")^2`
`2"KEm" = p^2`

Since we have calculated the momentum which is `1.0193846xx10^(-27)" kg m⋅ s"^(-2) ⋅ "s"` KE is

`2KE xx (9.11 xx 10^(-31) "kg") = 1.0193846e-27^2`

When the momentum squares the expression changes

`2KE xx (9.11 xx 10^(-31) "kg") = 1.039145xx10^(-54)" kg"^2 "m"^2 s^(-2) ⋅ s`

`2"KE" = (1.039145xx10^(-54)" kg"^cancel(2) "m"^2 s^(-2) ⋅ s)/(9.11 xx 10^(-31) cancel("kg"))`

`2KE = 1.1406641xx10^(-24)" kg m"^2 "s"^(-2)`

`KE = (1.1406641xx 10^(-24)"kg m"^2 "s"^(-2))/2`

`KE = 5.7033205xx10^(-25)" kg m"^2 "s"^(-2)`

Recall that `"kg m"^2 "s"^(-2)` is equat to joules

`KE = 5.7033205xx10^(-25) J`

It make's sense since the wavelength is so large .

Answer 2

`K = 5.7 xx 10^(-25)` `"J"`

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Consider the properties of electrons... Since electrons have mass `m` and a wavelength `lambda`, they follow the de Broglie relation:

`lambda = h/p = h/(mv)` `" "bb((1))`

where `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant and `p = mv` is the linear momentum.

The kinetic energy can be expressed as a function of the momentum:

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

Therefore, if we solve `(1)` for momentum, we can get the kinetic energy from `(2)`.

`p = h/lambda`

`=> K = (h"/"lambda)^2/(2m)`

`= ((6.626 xx 10^(-34) "kg"cdot"m"^(cancel(2))"/s")/(650 cancel"nm" xx cancel"1 m"/(10^9 cancel"nm")))^2xx1/(2*9.109 xx 10^(-31) cancel"kg")`

Since the units work out as `"kg"cdot"m"^2"/s"^2`, or `"N"cdot"m"`, or `"J"`, we have units of energy. Therefore, the kinetic energy is:

`color(blue)(K = 5.7 xx 10^(-25))` `color(blue)("J")`

to two sig figs.

It makes sense that the kinetic energy is very small; after all, the electron is very light...