Question

An atom emits a photon of wavelength 1.08 meters. What is the energy change occurring in the atom due to this emission? (Planck's constant is 6.626 × 10-34 joule seconds, the speed of light is 2.998 × 108 m/s)

Answer 1

The answer is 1.84×10^(-25).

Answer 2

Answer : The energy change occurring in the atom due to this emission is, $18.38\times 10^{-26}J$

Solution :

Formula used :

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

where,

$\Delta E$ = change in energy of photon = ?

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $2.998\times 10^8m/s$

$\lambda$ = wavelength = 1.08 m

Now put all the given values in the above formula, we get the energy of the photons.

$\Delta E=\frac{(6.626\times 10^{-34}Js)\times (2.998\times 10^8m/s)}{1.08m}$

$\Delta E=18.39\times 10^{-26}J$

Therefore, the energy change occurring in the atom due to this emission is, $18.38\times 10^{-26}J$