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…

Question

asked Jan 28, 2020 194k views

1 Answer

Coolsv · Answer 1 · 2020-02-02T14:08:55+0000

Answer:

$\lambda=540.16\ nm$

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,

$\lambda=540.16\ nm$