
Find energy of each of the photons which
1.Correspond to light of frequency \[3\times {{10}^{15}}Hz\].
2.Have wavelength of \[0.50{{A}^{o}}\]
Answer
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Hint: The energy which has been delivered by the combination of atoms has been passed from the core of the sun by its light particles and warmth is called photons. The photon is considered to be massless with no electric charge and the particle is stable. The photons usually have two possible polarization states.
Complete step-by-step answer:
The recurrence is considered as the quantity of waves that pass a point in the space during any of the time stretches generally equal to 1 second. We measure it in the units of cycles every second or in hertz.
(i)So the relation between the energy of photons and frequency or the formula which relates the two is :
\[E=h\nu \]
Here E is the energy of photons, h is the Planck constant and v is the frequency.
The values given to us are v= \[3\times {{10}^{15}}Hz\] and h= \[6.626\times {{10}^{-34}}\]Js.
So substituting the values in the above formula we get,
E=\[6.626\times {{10}^{-34}}\times 3\times {{10}^{15}}=1.9878\times {{10}^{-18}}J\]
So the energy of the photon is \[1.9878\times {{10}^{-18}}J\].
(ii)We know that 1 \[{{A}^{o}}={{10}^{-10}}m\], so \[0.50{{A}^{o}}=0.50\times {{10}^{-10}}m\]
Now the relationship or the formula which relates the energy of photon with the wavelength is the following:
\[E=\dfrac{hc}{\lambda }\]
Here E is the energy of a photon, h is Planck constant, c is the speed of light and \[\lambda \] is the wavelength.
So the values given to us are h=\[6.626\times {{10}^{-34}}\], c=\[3\times {{10}^{8}}\] and \[\lambda \]=\[0.50{{A}^{o}}=0.50\times {{10}^{-10}}m\]
On substituting the values in the above formula we get,
E=\[\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{0.50\times {{10}^{-10}}m}=3.98\times {{10}^{-15}}s\]
So the energy of the photon is \[3.98\times {{10}^{-15}}s\].
Note:The photon energy relies on its recurrence that is how quick is the electric field and the attractive field squirm. The higher is the recurrence the more energy will be of the photon. When the recurrence of the photon increases then the frequency decreases and vice versa.
Complete step-by-step answer:
The recurrence is considered as the quantity of waves that pass a point in the space during any of the time stretches generally equal to 1 second. We measure it in the units of cycles every second or in hertz.
(i)So the relation between the energy of photons and frequency or the formula which relates the two is :
\[E=h\nu \]
Here E is the energy of photons, h is the Planck constant and v is the frequency.
The values given to us are v= \[3\times {{10}^{15}}Hz\] and h= \[6.626\times {{10}^{-34}}\]Js.
So substituting the values in the above formula we get,
E=\[6.626\times {{10}^{-34}}\times 3\times {{10}^{15}}=1.9878\times {{10}^{-18}}J\]
So the energy of the photon is \[1.9878\times {{10}^{-18}}J\].
(ii)We know that 1 \[{{A}^{o}}={{10}^{-10}}m\], so \[0.50{{A}^{o}}=0.50\times {{10}^{-10}}m\]
Now the relationship or the formula which relates the energy of photon with the wavelength is the following:
\[E=\dfrac{hc}{\lambda }\]
Here E is the energy of a photon, h is Planck constant, c is the speed of light and \[\lambda \] is the wavelength.
So the values given to us are h=\[6.626\times {{10}^{-34}}\], c=\[3\times {{10}^{8}}\] and \[\lambda \]=\[0.50{{A}^{o}}=0.50\times {{10}^{-10}}m\]
On substituting the values in the above formula we get,
E=\[\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{0.50\times {{10}^{-10}}m}=3.98\times {{10}^{-15}}s\]
So the energy of the photon is \[3.98\times {{10}^{-15}}s\].
Note:The photon energy relies on its recurrence that is how quick is the electric field and the attractive field squirm. The higher is the recurrence the more energy will be of the photon. When the recurrence of the photon increases then the frequency decreases and vice versa.
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Find energy of each of the photons which
1.Correspond to light of frequency \[3\times {{10}^{15}}Hz\].
2.Have wavelength of \[0.50{{A}^{o}}\]
1.Correspond to light of frequency \[3\times {{10}^{15}}Hz\].
2.Have wavelength of \[0.50{{A}^{o}}\]

Structure of atom class 11 Chemistry -NCERT EXERCISE 2.6 | Chemistry | Sumandeep Mann
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