Question

An intense emission line for a new element is observed at a wavelength of 675 nm. What is the energy of a single photon of this light?Energy = ___.

Answer 1

Answer:

$E=2.95\times 10^{-19}\ J$

Explanation:

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

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

Given, $\lambda=675\ nm=675\times 10^{-9}\ m$

Thus, applying values as:

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{675\times 10^{-9}}\ J$

$E=\frac{10^{-26}\times \:19.878}{10^{-9}\times \:675}\ J$

$E=2.95\times 10^{-19}\ J$