Question

$\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{}\mathbf{Hz}$An atom (not a hydrogen atom) absorbs a photon whose associated frequency is . By what amount does the energy of the atom increase?

Short answer

The energy of the atom is increased by $4.12\times {10}^{-19}\mathrm{J}$.

Step-by-step solution

The energy of the photon:

The expression of the energy of the photon is given by,

$\mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{hf}$ ….. (1)

Here, h is plank constant, and f is frequency of the photon.

Define the amount of increase in the energy of the atom:

Substitute $6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}$for h, and $6.2\times {10}^{14}\mathrm{Hz}$for f in equation (1).

$$\begin{gathered} ∆\mathrm{E}=\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(6.2\times {10}^{14}\mathrm{Hz}\right) \\ =4.12\times {10}^{-19}\mathrm{J} \end{gathered}$$

Hence, the energy of the atom is increased by $4.12\times {10}^{-19}\mathrm{J}$.