
Frequency of a photon having energy 66 eV is,
A. \[8 \times {10^{ - 15}}Hz\]
B. \[12 \times {10^{ - 15}}Hz\]
C. \[16 \times {10^{15}}Hz\]
D. None of these
Answer
164.4k+ views
Hint:The frequency of the photon is given. So, to find the energy carried by the photon we use the quantization of energy formula. The energy of the photon is proportional to the frequency of the photon.
Formula used:
\[E = h\nu \],
here $h$ is the Plank’s constant and E is the energy of the photon with frequency equals to \[\nu \].
\[c = \nu \lambda \]
Here 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:
It is given that the energy of the photon is 66 eV. We need to find the frequency of the photon with the given energy. We need to change the unit of energy from eV to joule. One electron-Volt is the energy used to accelerate an electron in a region of electric potential 1 volt.
So, \[1eV = 1.6 \times {10^{ - 19}}J\]
Hence, the energy of the photon is,
\[E = 66 \times \left( {1.6 \times {{10}^{ - 19}}} \right)J\]
\[\Rightarrow E = 1.06 \times {10^{ - 17}}J\]
Hence, the energy of the given photon is \[1.06 \times {10^{ - 17}}J\].
We got the energy of the photon and now we need to use the formula of energy of the photon to find the frequency of the photon,
Using the expression for the energy of the photon, we find,
\[E = h\nu \]
\[\Rightarrow \nu = \dfrac{E}{h}\]
Putting the values of Plank’s constant and the energy, we get
\[\nu = \dfrac{{\left( {1.06 \times {{10}^{ - 17}}} \right)}}{{\left( {6.626 \times {{10}^{ - 34}}} \right)}}Hz\]
\[\Rightarrow \nu = 1.59 \times {10^{16}}Hz\]
On rounding-off the value to two decimal places, we get the frequency of the photon as,
\[\nu = 1.6 \times {10^{16}}Hz\]
\[\Rightarrow \nu = \left( {1.6 \times 10} \right) \times {10^{15}}Hz\]
\[\therefore \nu = 16 \times {10^{15}}Hz\]
Hence, the frequency of the given photon is \[16 \times {10^{15}}Hz\].
Therefore, the correct option is C.
Note: The given unit of the energy is electron-volt and the frequency is given in Hertz. So, we must change the given energy unit to Joules from electron volt. Then apply the energy formula.
Formula used:
\[E = h\nu \],
here $h$ is the Plank’s constant and E is the energy of the photon with frequency equals to \[\nu \].
\[c = \nu \lambda \]
Here 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:
It is given that the energy of the photon is 66 eV. We need to find the frequency of the photon with the given energy. We need to change the unit of energy from eV to joule. One electron-Volt is the energy used to accelerate an electron in a region of electric potential 1 volt.
So, \[1eV = 1.6 \times {10^{ - 19}}J\]
Hence, the energy of the photon is,
\[E = 66 \times \left( {1.6 \times {{10}^{ - 19}}} \right)J\]
\[\Rightarrow E = 1.06 \times {10^{ - 17}}J\]
Hence, the energy of the given photon is \[1.06 \times {10^{ - 17}}J\].
We got the energy of the photon and now we need to use the formula of energy of the photon to find the frequency of the photon,
Using the expression for the energy of the photon, we find,
\[E = h\nu \]
\[\Rightarrow \nu = \dfrac{E}{h}\]
Putting the values of Plank’s constant and the energy, we get
\[\nu = \dfrac{{\left( {1.06 \times {{10}^{ - 17}}} \right)}}{{\left( {6.626 \times {{10}^{ - 34}}} \right)}}Hz\]
\[\Rightarrow \nu = 1.59 \times {10^{16}}Hz\]
On rounding-off the value to two decimal places, we get the frequency of the photon as,
\[\nu = 1.6 \times {10^{16}}Hz\]
\[\Rightarrow \nu = \left( {1.6 \times 10} \right) \times {10^{15}}Hz\]
\[\therefore \nu = 16 \times {10^{15}}Hz\]
Hence, the frequency of the given photon is \[16 \times {10^{15}}Hz\].
Therefore, the correct option is C.
Note: The given unit of the energy is electron-volt and the frequency is given in Hertz. So, we must change the given energy unit to Joules from electron volt. Then apply the energy formula.
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