Question

Q. Show that energy E of a photon having wavelength λ can be written as E= (1240ev/ λ)nm.

Answer 1

Answer on Question 58641, Physics, Quantum Mechanics

Question:

Show that energy $E$ of a photon having wavelength $\lambda$ can be written as

$$E(eV) = \frac{1240\,eV \cdot nm}{\lambda(nm)}.$$

Answer:

As we know, there is an inverse relationship between the energy of the photon and the wavelength of the light given by the equation:

$$E = \frac{hc}{\lambda},$$

here, $h = 6.626 \cdot 10^{-34}\,J \cdot s$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of the light.

Let's multiply $h$ by $c$, we get:

$$hc = 6.626 \cdot 10^{-34}\,J \cdot s \cdot 3 \cdot 10^8\, \frac{m}{c} = 1.9878 \cdot 10^{-25}\,J \cdot m$$

When we divide this expression by $\lambda$, we get the energy of the photon in joules $(J)$. But, when we dealing with "particles" such as photons or electrons, a commonly used unit of energy is the electronvolt $(eV)$ rather than the joule $(J)$. An electronvolt is the energy required to raise an electron through 1 volt, thus a photon with an energy of $1\,eV = 1.602 \cdot 10^{-19}\,J$.

Let's rewrite the above expression in terms of $eV$:

$$hc = 1.9878 \cdot 10^{-25}\,J \cdot m \cdot \frac{1\,eV}{1.602 \cdot 10^{-19}\,J} = 1240 \cdot 10^{-9}\,eV \cdot m.$$

Then, by the condition of the question, we need to have the units for $\lambda$ be in $nm$:

$$hc = 1240 \cdot 10^{-9}\,eV \cdot m \cdot \frac{10^9\,nm}{m} = 1240\,eV \cdot nm.$$

Finally, we get the energy of the photon in terms of $eV$ and $nm$, thus, the expression which relates the energy and wavelength of the photon in our case looks like:

$$E(eV) = \frac{hc}{\lambda(nm)} = \frac{1240 \, eV \cdot nm}{\lambda(nm)},$$

This formula can be useful in solving many problems. Let's, for example, consider the orange-colored light from a highway sodium lamp that has a wavelength of $589 \, nm$. How much energy is possessed by an individual photon from such a lamp? Using our formula, we get:

$$E(eV) = \frac{hc}{\lambda(nm)} = \frac{1240 \, eV \cdot nm}{\lambda(nm)} = \frac{1240 \, eV \cdot nm}{589 \, nm} = 2.105 \, eV.$$

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