Question

An atom emits of wavelength 1.08 meters. what is the energy change occurring in the Atom due to this emission? (Planck’s constant is 6.626x10^-34 joule seconds, the speed of light is 2.998x10^8m/s)

Answer 1

To answer this question, we can use the following equation:

$e=\frac{h*c}{lambda}$

where e is the energy change, h is Planck's constant, c is the speed of light (m/s), and lambda is wavelength in meters. So then we can plug in the information:

$e=\frac{(6.626x10^{-34}J*s)(2.998x10^{8}\frac{m}{s})}{1.08m}$

$e=\frac{1.9865x10^{-25}J}{1.08}$

$e=1.84x10^{-25}J$

So the energy that is omitted by the atom is $1.84x10^{-25}J$.