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

May 6, 2017

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

#### Explanation:

The De Broglie equation states that the

${\text{KE" = (1/2)"mv}}^{2}$

$\text{Where KE is kinetic energy}$
$\text{Where m is mass and is a constant for each particle}$
$\text{Where v is velocity}$

$2 {\text{KE" = "mv}}^{2}$

$2 {\text{KEm" = "m"^2"v}}^{2}$

$\text{mv" = "p}$

$\therefore {\text{2KEm" = "p}}^{2}$

$\therefore p = \sqrt{\text{2KEm}}$

Another equation of De Broglie's work is

λ = "h"/"p"

Where $\text{h}$ 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 $9.11 \times {10}^{- 31} \text{kg}$

Convert nm to m

650nm = $650 \times {10}^{-} 9 m$

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
$2 \text{KEm} = {p}^{2}$

Since we have calculated the momentum which is $1.0193846 \times {10}^{- 27} \text{ kg m⋅ s"^(-2) ⋅ "s}$ KE is

$2 K E \times \left(9.11 \times {10}^{- 31} \text{kg}\right) = {1.0193846e-27}^{2}$

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"))

$2 K E = 1.1406641 \times {10}^{- 24} {\text{ kg m"^2 "s}}^{- 2}$

$K E = \frac{1.1406641 \times {10}^{- 24} {\text{kg m"^2 "s}}^{- 2}}{2}$

$K E = 5.7033205 \times {10}^{- 25} {\text{ kg m"^2 "s}}^{- 2}$

Recall that ${\text{kg m"^2 "s}}^{- 2}$ is equat to joules

$K E = 5.7033205 \times {10}^{- 25} J$

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

May 6, 2017

$K = 5.7 \times {10}^{- 25}$ $\text{J}$

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

$\lambda = \frac{h}{p} = \frac{h}{m v}$ $\text{ } \boldsymbol{\left(1\right)}$

where $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant and $p = m v$ is the linear momentum.

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

$K = \frac{1}{2} m {v}^{2} = {p}^{2} / \left(2 m\right)$$\text{ } \boldsymbol{\left(2\right)}$

Therefore, if we solve $\left(1\right)$ for momentum, we can get the kinetic energy from $\left(2\right)$.

$p = \frac{h}{\lambda}$

$\implies K = {\left(h \text{/} \lambda\right)}^{2} / \left(2 m\right)$

= ((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 ${\text{kg"cdot"m"^2"/s}}^{2}$, or $\text{N"cdot"m}$, or $\text{J}$, we have units of energy. Therefore, the kinetic energy is:

$\textcolor{b l u e}{K = 5.7 \times {10}^{- 25}}$ $\textcolor{b l u e}{\text{J}}$

to two sig figs.

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