Calculate the number of photons emitted per second by a 10 W sodium vapour lamp. Assume that 60% of the consumed energy is converted into light. Wavelength of sodium light = 590 nm.
Answer
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Hint: In the question, they’ve asked us to find the number of photons emitted per second by a 10 W sodium vapour lamp. In order to find it, we first need to calculate the energy that is converted into light. Then, we can calculate the number of photons based on the relation between energy and wavelength of the light incident.
Formula used:
$E = nh\dfrac{c}{\lambda }$
Complete answer:
They’ve given that a sodium vapour lamp of 10 W converts 60% of the consumed energy into light.
The power is the energy drawn per second. So, 10 J of energy is drawn in one second. The energy that is converted into light will be
$10J \times \dfrac{{60}}{{100}} = 6J$
So, 6 J of energy is converted into light energy.
The energy emitted by photons is given by
$E = nh\dfrac{c}{\lambda }$
Where,
$E$ is the energy emitted by the photons
$n$ is the no. of photons emitted
$h$ is the Planck’s constant given by $6.63 \times {10^{ - 34}}Js$
$c$ is the speed of the light given by $3 \times {10^8}m{s^{ - 1}}$
$\lambda $ is the wavelength of the incident light
From the data given in the question, we have the wavelength of light to be $\lambda = 590nm$, the energy of the emitted photons will be the energy converted into the light which is $E = 6J$. Substituting these quantities in the formula we will have
$\eqalign{
& E = nh\dfrac{c}{\lambda } \cr
& \Rightarrow n = \dfrac{{E\lambda }}{{hc}} \cr
& \Rightarrow n = \dfrac{{6 \times 590 \times {{10}^{ - 9}}}}{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \cr
& \Rightarrow n = 1.78 \times {10^{19}} \cr} $
Therefore, $1.78 \times {10^{19}}$ the number of photons will be released per second by the sodium vapour lamp.
Note:
In this problem, we consider the particle nature of light. According to the particle nature of light, it is made of packets of energy or particles called photons. These photons have fixed energy for a given wavelength. Always pay attention to the suffixes or factors given in the units. It is also better to remember constants like c and h, as they will be used in a wide range of problems. And the number of photons do not have any units.
Formula used:
$E = nh\dfrac{c}{\lambda }$
Complete answer:
They’ve given that a sodium vapour lamp of 10 W converts 60% of the consumed energy into light.
The power is the energy drawn per second. So, 10 J of energy is drawn in one second. The energy that is converted into light will be
$10J \times \dfrac{{60}}{{100}} = 6J$
So, 6 J of energy is converted into light energy.
The energy emitted by photons is given by
$E = nh\dfrac{c}{\lambda }$
Where,
$E$ is the energy emitted by the photons
$n$ is the no. of photons emitted
$h$ is the Planck’s constant given by $6.63 \times {10^{ - 34}}Js$
$c$ is the speed of the light given by $3 \times {10^8}m{s^{ - 1}}$
$\lambda $ is the wavelength of the incident light
From the data given in the question, we have the wavelength of light to be $\lambda = 590nm$, the energy of the emitted photons will be the energy converted into the light which is $E = 6J$. Substituting these quantities in the formula we will have
$\eqalign{
& E = nh\dfrac{c}{\lambda } \cr
& \Rightarrow n = \dfrac{{E\lambda }}{{hc}} \cr
& \Rightarrow n = \dfrac{{6 \times 590 \times {{10}^{ - 9}}}}{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \cr
& \Rightarrow n = 1.78 \times {10^{19}} \cr} $
Therefore, $1.78 \times {10^{19}}$ the number of photons will be released per second by the sodium vapour lamp.
Note:
In this problem, we consider the particle nature of light. According to the particle nature of light, it is made of packets of energy or particles called photons. These photons have fixed energy for a given wavelength. Always pay attention to the suffixes or factors given in the units. It is also better to remember constants like c and h, as they will be used in a wide range of problems. And the number of photons do not have any units.
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