Question

Calculate the wavelength, wavenumber and frequency of photons having an energy equal to three electron volt. $\left( {h = 6 \cdot 62 \times {{10}^{ - 27}}{\text{erg - s}}} \right)$.

Answer 1

Hint: Photons are the particles which have electromagnetic radiation. Wavelength is the consecutive difference between two crests or troughs in a wave. The formula of the energy of the wave can be used to calculate the frequency of the wave The relation between the speed, frequency and wavelength can be used to calculate the wavelength and wave number.

Formula used: The energy of the wave is given by $E = h \cdot v$ where $E$ is the energy $h$ is the Planck’s constant and $v$ is the frequency of the wave.

Complete step by step answer:
It is given that the energy of the photons is equal to 3eV (electron volts).
As it is given that the energy of the wave is given by $E = h \cdot v$ where $E$ is the energy $h$ is the Planck’s constant and $v$ is the frequency of the wave.
Therefore,
$ \Rightarrow E = h \cdot v$
$ \Rightarrow v = \dfrac{E}{h}$
$ \Rightarrow v = \dfrac{{3 \cdot \left( {1 \cdot 602 \times {{10}^{ - 12}}} \right)}}{{6 \cdot 62 \times {{10}^{ - 27}}}}$ (As the energy is equal to 3eV and $1e{\text{V = 1}} \cdot {\text{602}} \times {\text{1}}{{\text{0}}^{ - 12}}erg$).
$ \Rightarrow v = 7 \cdot 26 \times {10^{14}}{s^{ - 1}}$
$ \Rightarrow v = 7 \cdot 26 \times {10^{14}}Hz$
Since, $\lambda = \dfrac{c}{v}$ where $\lambda $ is wavelength, $c$ is speed of light and $v$ is the frequency of the wave.
$ \Rightarrow \lambda = \dfrac{{3 \times {{10}^8}}}{{7 \cdot 26 \times {{10}^{14}}Hz}}$
$ \Rightarrow \lambda = 4 \cdot 132 \times {10^{ - 5}}cm$
So the wavelength of the wave is $\lambda = 4 \cdot 132 \times {10^{ - 5}}cm$.
Also let us calculate the wave number. As the wave number is given by $\bar v = \dfrac{1}{\lambda }$ where $\bar v$ is the wave number and $\lambda $ is wavelength.
$ \Rightarrow \bar v = \dfrac{1}{\lambda }$
$ \Rightarrow \bar v = \dfrac{1}{{4 \cdot 132 \times {{10}^{ - 5}}cm}}$
$ \Rightarrow \bar v = 2 \cdot 42 \times {10^4}c{m^{ - 1}}$
So the wave number is given by $\bar v = 2 \cdot 42 \times {10^4}c{m^{ - 1}}$.

Therefore, the wavelength of the wave is $\lambda = 4 \cdot 132 \times {10^{ - 5}}cm$, the frequency of the wave is $v = 7 \cdot 26 \times {10^{14}}Hz$ and the wave number of the wave is given by $\bar v = 2 \cdot 42 \times {10^4}c{m^{ - 1}}$ given that the energy of the photons is three electron volts.

Note:
It is important for students to remember the formula of energy of the wave as it can help in solving these types of problems. The value of 1eV is equal to $1e{\text{V = 1}} \cdot {\text{602}} \times {\text{1}}{{\text{0}}^{ - 12}}{\text{erg}}$. The energy of the photon is given as 3eV.