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* 10^(-26)J`

Solution :

Formula used :

`\Delta E=(h* c)/(\lambda)`

where,

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

h = Planck's constant =
`6.626* 10^(-34)Js`

c = speed of light =
`2.998* 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=((6.626* 10^(-34)Js)* (2.998* 10^8m/s))/(1.08m)`

`\Delta E=18.39* 10^(-26)J`

Therefore, the energy change occurring in the atom due to this emission is,
`18.38* 10^(-26)J`