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

Answer:

$\lambda=540.16\ nm$

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 = $2.3\times 1.60\times 10^{-19}\ J=3.68\times 10^{-19}\ J$

Considering

$Energy=\frac {h\times c}{\lambda}$

Where,  

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

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

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

Thus,  

$3.68\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\frac{3.68}{10^{19}}=\frac{19.878}{10^{26}\lambda}$

$3.68\times \:10^{26}\lambda=1.9878\times 10^{20}$

$\lambda=5.40163\times 10^{-7}\ m=540.16\times 10^{-9}\ m$

Also,  

1 m = 10⁻⁹ nm

So,  

$\lambda=540.16\ nm$