
A laser used to read CDs emits red light of wavelength 700 nanometers. How many photons does it emit each second if its power is 0.1 w.
Answer
524.4k+ views
Hint: We have to use the following formula to calculate the energy of one photon.
E = h$v$
Here E = Energy of one photon
h = Planck’s constant
$v$ = frequency of the photon.
Complete answer:
- In the question it is asked to calculate how many photons are going to be released per second if the emitted light has a wavelength of 700 nm if the power is 0.1 w.
- The formula used to calculate the energy of the photon is as follows.
\[\begin{align}
& E\text{ }=\text{ }h~v \\
& E=h\times \dfrac{c}{\lambda } \\
\end{align}\]
Here E = Energy of one photon
h = Planck’s constant = $6.6023\times {{10}^{-34}}J/\sec $
c = velocity of the photon $3\times {{10}^{8}}m/s$
$\lambda $ = wavelength of the light = 72 nm = $700\times {{10}^{-9}}m$
- By substituting all the known values in the above to get the energy of the photon whose wavelength is 720 nm.
\[\begin{align}
& E=h\times \dfrac{c}{\lambda } \\
& E=\dfrac{6.6023\times {{10}^{-34}}\times 3.0\times {{10}^{8}}}{700\times {{10}^{-9}}} \\
& E=2.81\times {{10}^{-19}}J \\
\end{align}\]
- We know that 1 watt is nothing but 1 joule of work is done.
- That means 0.1 joule of work is done in one second.
- So $2.81\times {{10}^{-19}}J$ work done by one photon.
- Therefore one joule is going to produce $\dfrac{1}{2.81\times {{10}^{-19}}J}$ photons.
- Then we have to calculate 0.1 joule is going to be produced by how many photons.
\[\dfrac{1}{2.81\times {{10}^{-19}}J}\times 0.1=3.5\times {{10}^{17}}photons\]
- Means $3.5\times {{10}^{17}}photons$ are going to release in one second.
Note:
First we have to calculate the energy of the photons which are going to release from 720 nm of wavelength. Later we have to convert the energy into a number of protons with a power of 0.1 watt.
E = h$v$
Here E = Energy of one photon
h = Planck’s constant
$v$ = frequency of the photon.
Complete answer:
- In the question it is asked to calculate how many photons are going to be released per second if the emitted light has a wavelength of 700 nm if the power is 0.1 w.
- The formula used to calculate the energy of the photon is as follows.
\[\begin{align}
& E\text{ }=\text{ }h~v \\
& E=h\times \dfrac{c}{\lambda } \\
\end{align}\]
Here E = Energy of one photon
h = Planck’s constant = $6.6023\times {{10}^{-34}}J/\sec $
c = velocity of the photon $3\times {{10}^{8}}m/s$
$\lambda $ = wavelength of the light = 72 nm = $700\times {{10}^{-9}}m$
- By substituting all the known values in the above to get the energy of the photon whose wavelength is 720 nm.
\[\begin{align}
& E=h\times \dfrac{c}{\lambda } \\
& E=\dfrac{6.6023\times {{10}^{-34}}\times 3.0\times {{10}^{8}}}{700\times {{10}^{-9}}} \\
& E=2.81\times {{10}^{-19}}J \\
\end{align}\]
- We know that 1 watt is nothing but 1 joule of work is done.
- That means 0.1 joule of work is done in one second.
- So $2.81\times {{10}^{-19}}J$ work done by one photon.
- Therefore one joule is going to produce $\dfrac{1}{2.81\times {{10}^{-19}}J}$ photons.
- Then we have to calculate 0.1 joule is going to be produced by how many photons.
\[\dfrac{1}{2.81\times {{10}^{-19}}J}\times 0.1=3.5\times {{10}^{17}}photons\]
- Means $3.5\times {{10}^{17}}photons$ are going to release in one second.
Note:
First we have to calculate the energy of the photons which are going to release from 720 nm of wavelength. Later we have to convert the energy into a number of protons with a power of 0.1 watt.
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