Question

What is the energy of an electromagnetic wave that has a wavelength of 7.0×10^-12 m? Use the equation E- Intion hc

Answer 1

Answer:

Assume that this wavelength is measured in vacuum. The energy on each photon of this wave would be approximately $2.8 \times 10^{-14}\; \rm J$.

Explanation:

The Planck-Einstein Relation relates the energy $E$ of a photon to its frequency $f$:

$E = h \, f$,

where $h$ is Planck's Constant.

$h \approx 6.67 \times 10^{-34}\; \rm m^2\cdot kg \cdot s^{-1}$.

This question did not provide the frequency $f$ of this wave directly; the value of $f$ needs to be calculated from the wavelength $\lambda$ of this wave. Assume that this wave is travelling at the speed of light in vacuum:

$c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}$.

The frequency of this electromagnetic wave would be:

$\begin{aligned}f &= \frac{c}{\lambda}\approx \frac{3.00 \times 10^{8}\; \rm m \cdot s^{-1}}{7.0 \times 10^{-12}\; \rm m} \\ &\approx 4.29\times 10^{19}\; \rm s^{-1} = 4.29 \times 10^{19}\; \rm Hz\end{aligned}$.

Apply the Planck-Einstein Relation to find the energy of a photon of this electromagnetic wave:

$\begin{aligned}E &= h \, f \\ &\approx 6.63 \times 10^{-34}\; \rm m^2\cdot kg \cdot s^{-1} \times 4.29 \times 10^{19}\; \rm s^{-1} \\ &\approx 2.8 \times 10^{-14}\; \rm m^2\cdot kg \cdot s^{-1} = 2.8 \times 10^{-14}\; \rm J \end{aligned}$.

Note that combining the two equations above ($E = h \, f$ and $\displaystyle f = \frac{c}{\lambda}$) will give:

$\displaystyle E = h\cdot \frac{c}{\lambda}$.

This equation is supposed to give the same result (energy of a photon of this wave given its wavelength and speed) in one step:

$\begin{aligned}E &= h\cdot \frac{c}{\lambda} \\ &\approx 6.63 \times 10^{-34} \; \rm m^2\cdot kg \cdot s^{-1}\times \frac{3.00\times 10^{8}\; \rm m \cdot s^{-1}}{7.0 \times 10^{-12}\; \rm m} \\ &\approx 2.8 \times 10^{-14}\; \rm m\end{aligned}$.