Answer
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Hint: Wavenumber is defined as the number of waves per unit time. It can be expressed in terms of wavelength where wavelength can be determined from Planck’s constant, velocity of light, and given energy in the below formula. From the obtained wavelength, the wave number can be calculated.
Complete answer:
The equation for the energy can be written as $ E = \dfrac{{hc}}{\lambda } $
Where $ E $ is the energy given in joules.
$ h $ is Planck’s constant which has the value of $ 6.626 \times {10^{ - 34}}J.s $
$ c $ is the velocity of light which has the value of $ 3 \times {10^8}m{s^{ - 1}} $
$ \lambda $ is the wavelength
Thus, the wavelength can be expressed as $ \lambda = \dfrac{{hc}}{E} $
By substituting the above values in the formula of wavelength
$ \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}J.s \times 3 \times {{10}^8}m{s^{ - 1}}}}{{6.626 \times {{10}^{ - 20}}J}} $
The value of $ \lambda $ will be $ 3 \times {10^{ - 6}}m $
The wavenumber has the formula of $ k = \dfrac{{2\pi }}{\lambda } $
Substitute the obtained wavelength in the above equation,
$ k = \dfrac{{2 \times 3.14}}{{3 \times {{10}^{ - 6}}m}} $
The value of wavenumber will be $ 2.093 \times {10^{ - 6}}waves{\left( m \right)^{ - 1}} $
Thus, the wavenumber of photon whose energy is $ 6.626 \times {10^{ - 20}}J $ will be $ 2.093 \times {10^{ - 6}}waves{\left( m \right)^{ - 1}} $
Wavenumber can also be expressed in different terms in spectroscopic chemistry by the equation $ \overline \vartheta = \dfrac{E}{{hc}} $
Where $ \overline \vartheta $ is wavenumber
$ E $ is energy
By substituting the value of velocity of light, Planck’s constant, and energy in the above formula.
$ \overline \vartheta = \dfrac{{6.626 \times {{10}^{ - 20}}J}}{{6.626 \times {{10}^{ - 34}}J.s \times 3 \times {{10}^8}m{s^{ - 1}}}} $
By further simplification, the wavenumber will be $ 3.336 \times {10^3}c{m^{ - 1}} $
Thus, the wavenumber of a photon will be $ 3.336 \times {10^3}c{m^{ - 1}} $
Note:
In physics, the formula of wave number can be expressed as the first formula. In spectroscopic chemistry, the wave number is the reciprocal of wavelength. Thus, the two formulas were different which gives the different values. Wavenumber was expressed as $ k $ and $ \overline \vartheta $ .
Complete answer:
The equation for the energy can be written as $ E = \dfrac{{hc}}{\lambda } $
Where $ E $ is the energy given in joules.
$ h $ is Planck’s constant which has the value of $ 6.626 \times {10^{ - 34}}J.s $
$ c $ is the velocity of light which has the value of $ 3 \times {10^8}m{s^{ - 1}} $
$ \lambda $ is the wavelength
Thus, the wavelength can be expressed as $ \lambda = \dfrac{{hc}}{E} $
By substituting the above values in the formula of wavelength
$ \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}J.s \times 3 \times {{10}^8}m{s^{ - 1}}}}{{6.626 \times {{10}^{ - 20}}J}} $
The value of $ \lambda $ will be $ 3 \times {10^{ - 6}}m $
The wavenumber has the formula of $ k = \dfrac{{2\pi }}{\lambda } $
Substitute the obtained wavelength in the above equation,
$ k = \dfrac{{2 \times 3.14}}{{3 \times {{10}^{ - 6}}m}} $
The value of wavenumber will be $ 2.093 \times {10^{ - 6}}waves{\left( m \right)^{ - 1}} $
Thus, the wavenumber of photon whose energy is $ 6.626 \times {10^{ - 20}}J $ will be $ 2.093 \times {10^{ - 6}}waves{\left( m \right)^{ - 1}} $
Wavenumber can also be expressed in different terms in spectroscopic chemistry by the equation $ \overline \vartheta = \dfrac{E}{{hc}} $
Where $ \overline \vartheta $ is wavenumber
$ E $ is energy
By substituting the value of velocity of light, Planck’s constant, and energy in the above formula.
$ \overline \vartheta = \dfrac{{6.626 \times {{10}^{ - 20}}J}}{{6.626 \times {{10}^{ - 34}}J.s \times 3 \times {{10}^8}m{s^{ - 1}}}} $
By further simplification, the wavenumber will be $ 3.336 \times {10^3}c{m^{ - 1}} $
Thus, the wavenumber of a photon will be $ 3.336 \times {10^3}c{m^{ - 1}} $
Note:
In physics, the formula of wave number can be expressed as the first formula. In spectroscopic chemistry, the wave number is the reciprocal of wavelength. Thus, the two formulas were different which gives the different values. Wavenumber was expressed as $ k $ and $ \overline \vartheta $ .
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