# Question 3da69

Feb 21, 2017

$\textsf{\lambda = 1.23 \textcolor{w h i t e}{x} n m}$

#### Explanation:

1 electron volt = $\textsf{1.6 \times {10}^{- 19} \textcolor{w h i t e}{x} J}$

We can equate this to the kinetic energy of the electron.

$\therefore$$\textsf{\frac{1}{2} m {v}^{2} = 1.6 \times {10}^{- 19} \textcolor{w h i t e}{x} J}$

The mass of the electron (ignoring relativistic effects) =$\textsf{9.1 \times {10}^{- 31} \textcolor{w h i t e}{x} \text{kg}}$

$\therefore$$\textsf{{v}^{2} = \frac{2 \times 1.6 \times {10}^{- 19}}{9.1 \times {10}^{- 31}} = 0.35 \times {10}^{12}}$

$\textsf{v = \sqrt{0.35 \times {10}^{12}} = 5.91 \times {10}^{5} \textcolor{w h i t e}{x} \text{m/s}}$

Now we use the de Broglie expression:

sf(lambda=h/(mv)#

$\therefore$$\textsf{\lambda = \frac{6.63 \times {10}^{- 34}}{9.1 \times {10}^{- 31} \times 5.91 \times {10}^{5}} \textcolor{w h i t e}{x} m}$

$\textsf{\lambda = 1.23 \times {10}^{- 9} \textcolor{w h i t e}{x} m}$

$\textsf{\lambda = 1.23 \textcolor{w h i t e}{x} n m}$