
Energy of a photon of light of wavelength 450 nm is
A. \[4.4 \times {10^{ - 19}}J\]
B. \[2.5 \times {10^{ - 19}}J\]
C. \[1.25 \times {10^{ - 17}}J\]
D. \[2.5 \times {10^{ - 17}}J\]
Answer
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Hint: The photon is the light wave's qualitative energy unit. It varies with the frequency of the light pulse. The frequency of the light wave is the most distinguishing quality since it is unaffected by changes in the medium through which the light travels.
Formula used:
\[E = h\nu \],
where h is the Plank’s constant and E is the energy of the photon with a frequency equal to \[\nu \].
\[c = \nu \lambda \],
where c is the speed of light, \[\nu \] is the frequency of the photon and \[\lambda \] is the wavelength of the light wave.
Complete step by step solution:
The wavelength of the photon is given as 450 nm.
As we know that \[1nm = {10^{ - 9}}m\]
The wavelength of the given photon in S.I. unit will be,
\[\lambda = 450 \times {10^{ - 9}}m\]
\[\Rightarrow \lambda = 4.50 \times {10^{ - 7}}m\]
We got the wavelength of the photon and now we need to use the relation of the characteristic physical quantities for the electromagnetic wave to find the frequency of the photon,
\[c = \nu \lambda \]
\[\Rightarrow \nu = \dfrac{c}{\lambda }\]
On putting the value of the speed of light and the wavelength of the photon, we get
\[\nu = \dfrac{{3 \times {{10}^8}}}{{4.50 \times {{10}^{ - 9}}}}Hz\]
\[\Rightarrow \nu = 6.67 \times {10^{16}}Hz\]
Using the expression for the energy of the photon, we find,
\[E = h\nu \]
Putting the values of Plank’s constant and the frequency, we get
\[E = \left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {6.67 \times {{10}^{16}}} \right)J\]
\[\therefore E = 4.4 \times {10^{ - 17}}J\]
Hence, the energy of the given photon is \[4.4 \times {10^{ - 17}}J\].
Therefore, the correct option is A.
Note: When addressing numerical problems, we must be careful of the physical quantity's units. We must convert all of the data provided into standard unit form. The energy of a photon is given either in Joules or in electron-Volts. One electron-Volt is the amount of energy that an electron has when it is accelerated across a potential difference of one volt.
Formula used:
\[E = h\nu \],
where h is the Plank’s constant and E is the energy of the photon with a frequency equal to \[\nu \].
\[c = \nu \lambda \],
where c is the speed of light, \[\nu \] is the frequency of the photon and \[\lambda \] is the wavelength of the light wave.
Complete step by step solution:
The wavelength of the photon is given as 450 nm.
As we know that \[1nm = {10^{ - 9}}m\]
The wavelength of the given photon in S.I. unit will be,
\[\lambda = 450 \times {10^{ - 9}}m\]
\[\Rightarrow \lambda = 4.50 \times {10^{ - 7}}m\]
We got the wavelength of the photon and now we need to use the relation of the characteristic physical quantities for the electromagnetic wave to find the frequency of the photon,
\[c = \nu \lambda \]
\[\Rightarrow \nu = \dfrac{c}{\lambda }\]
On putting the value of the speed of light and the wavelength of the photon, we get
\[\nu = \dfrac{{3 \times {{10}^8}}}{{4.50 \times {{10}^{ - 9}}}}Hz\]
\[\Rightarrow \nu = 6.67 \times {10^{16}}Hz\]
Using the expression for the energy of the photon, we find,
\[E = h\nu \]
Putting the values of Plank’s constant and the frequency, we get
\[E = \left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {6.67 \times {{10}^{16}}} \right)J\]
\[\therefore E = 4.4 \times {10^{ - 17}}J\]
Hence, the energy of the given photon is \[4.4 \times {10^{ - 17}}J\].
Therefore, the correct option is A.
Note: When addressing numerical problems, we must be careful of the physical quantity's units. We must convert all of the data provided into standard unit form. The energy of a photon is given either in Joules or in electron-Volts. One electron-Volt is the amount of energy that an electron has when it is accelerated across a potential difference of one volt.
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