Question

What is the wavelength in picometers for an electron with 360 eV of kinetic energy?

Answer 1

Wavelength = 64.635 pm

Step-by-step explanation:

The expression for the deBroglie wavelength and kinetic energy is:

`\lambda=\frac {h}{√(2* m* K.E.)}`

Where,

`\lambda` is the deBroglie wavelength

h is Plank's constant having value
`6.626* 10^(-34)\ Js`

m is the mass of electron having value
`9.11* 10^(-31)\ kg`

K.E. is the kinetic energy of the electron.

Given, K.E. = 360 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, K.E. =
`360* 1.6022* 10^(-19)\ J=5.76792* 10^(-17)\ J`

Applying in the equation as:

`\lambda=\frac {h}{√(2* m* K.E.)}`

`\lambda=\frac{6.626* 10^(-34)}{\sqrt {{2* 9.11* 10^(-31)* 5.76792* 10^(-17)}}}\ m`

`\lambda=\frac{10^(-34)* \:6.626}{\sqrt{10^(-48)* \:105.0915024}}\ m`

`\lambda=6.4635* 10^(-11)\ m`

Also, 1 m = 10¹² pm

So, Wavelength = 64.635 pm