Question

The energy in eV for light with a wavelength of 6250 angstroms is _. Note - there are 1.6 x 10-12 erg in 1 eV.

Answer 1

Answer:

2 eV

Explanation:

The energy of a photon of light is given by the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

In this problem we have:

$h=6.63\cdot 10^{-34} Js$

$c=3.0\cdot 10^8 m/s$

$\lambda=6250 A = 6250\cdot 10^{-10} m$ is the wavelength of the photon

Therefore, the energy in Joules is

$E=\frac{(6.63\cdot 10^{-34})(3.0\cdot 10^8)}{6250\cdot 10^{-10}}=3.2\cdot 10^{-19}J$

We want to convert this energy into electronvolts: we know that the conversion factor is

$1 eV = 1.6\cdot 10^{-19}J$

Therefore,

$E=\frac{3.2\cdot 10^{-19}}{1.6\cdot 10^{-19}}=2 eV$