Question

Determine the energy in joules and electron volts of a quantum with a wavelength of 600.0nm

Answer 1

The energy of quantum is given as follows:

$$E = h\nu$$

where $h \approx 6.626\times 10^{-34}J\cdot s$ or $h = 4.136\times 10^{-15}eV\cdot s$ is the Plank's constant, and

$$\nu = \dfrac{c}{\lambda}$$

is the frequency of the quantum, where $c = 3\times10^8m/s$ is the speed of light, and $\lambda = 600nm = 6\times 10^{-7}m$ is the wavelength.

Thus, obtain for joules:

$$E = \dfrac{hc}{\lambda} = \dfrac{6.626\times 10^{-34}\cdot 3\times 10^8}{6\times 10^{-7}} \approx 3.313\times10^{-19}J$$

In electron volts:

$$E = \dfrac{hc}{\lambda} = \dfrac{4.136\times 10^{-15}\cdot 3\times 10^8}{6\times 10^{-7}} \approx 2.068eV$$

Answer. $3.313\times10^{-19}J$ or $2.068eV$.