Chapter 27.1, Problem 1PP

To determine

To Find: Energy of photon.

Answer to Problem 1PP

  $\text{E =3}\text{.86}\times {\text{10}}^{-19}\text{J}$

Explanation of Solution

Given data:

Wavelength of photon, λ= 515 nm

Formula used:

Energy of the photon can be obtained by:

  $E=\frac{1240\text{eV}\cdot \text{nm}}{\lambda }$

Where, $\lambda$ is the wavelength.

Calculation:

  $\begin{array}{l}E=\frac{1240\text{eV}\cdot \text{nm}}{\lambda }\\ E=\frac{1240\text{eV}\cdot \text{nm}}{515\text{nm}}=2.41\text{eV}\\ \text{1}\text{eV}=1.6\times {10}^{-19}\text{J}\\ E=\text{2}\text{.41}\times 1.6\times {10}^{-19}\text{J}\\ E=\text{3}\text{.86}\times {\text{10}}^{-19}\text{J}\end{array}$

Conclusion:

The energy of the photon is $\text{E =3}\text{.86}\times {\text{10}}^{-19}\text{J}$ .