Question

A photon with 2.3 eV of energy can eject an electron from potassium. What is the corresponding wavelength of this type of light? Please answer in nm. 1 eV = 1.60 x 10-19 J Speed of light = 3.0 x 108 m/s Planck's constant = 6.626 x 10-34 Js socratic.

Answer 1

`\lambda=540.16\ nm`

Step-by-step explanation:

Given that:

The energy of the photon = 2.3 eV

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

1 eV = 1.60 × 10⁻¹⁹ J

So, Energy =
`2.3* 1.60* 10^(-19)\ J=3.68* 10^(-19)\ J`

Considering

`Energy=\frac {h* c}{\lambda}`

Where,

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

c is the speed of light having value
`3* 10^8\ m/s`

`\lambda` is the wavelength of the light being bombarded

Thus,

`3.68* 10^(-19)=\frac {6.626* 10^(-34)* 3* 10^8}{\lambda}`

`(3.68)/(10^(19))=(19.878)/(10^(26)\lambda)`

`3.68* \:10^(26)\lambda=1.9878* 10^(20)`

`\lambda=5.40163* 10^(-7)\ m=540.16* 10^(-9)\ m`

Also,

1 m = 10⁻⁹ nm

So,

`\lambda=540.16\ nm`