
If ${n_R}$ and ${n_V}$ denote the number of photons emitted by a red bulb and violet bulb of equal power in a given time, then:
A) ${n_R} = {n_V}$
B) ${n_R} > {n_V}$
C) ${n_R} < {n_V}$
D) ${n_R} \geqslant {n_V}$
Answer
219k+ views
Hint: In this question, we will use the concept of the Planck's energy equation that is the relation of the Planck's constant, the wavelength of the light, and the speed of the light. Then we will consider the number of photons required to achieve that much amount of energy.
Complete solution:
In this question, It is given that ${n_R}$ and ${n_V}$ denote the number of photons emitted by a red bulb and violet bulb of equal power in a given time. We need to find the relation between the number of photons emitted by a red bulb and by the violet bulb.
We will use the Planck law to get the energy as,
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$
Here, the planck's constant is $h$, the speed of light is $c$, and the wavelength of the light is $\lambda $.
As we know that photons are the light whose move in packets and bundles.
If the power of each photon is $P$, then energy given out in t second is equal to $pt$. Let the number of photons be $n$, then
$ \Rightarrow n = \dfrac{{Pt}}{E}$
Now we will substitute the values in the above relation.
\[ \Rightarrow n = \dfrac{{Pt}}{{\left( {hc/\lambda } \right)}}\]
Now we simplify the above expression.
\[ \Rightarrow n = \dfrac{{Pt\lambda }}{{hc}}\]
For red light,
$ \Rightarrow {n_R} = \dfrac{{Pt{\lambda _R}}}{{hc}}$
For violet light,
$ \Rightarrow {n_V} = \dfrac{{Pt{\lambda _V}}}{{hc}}$
Since, from above relation we get,
$ \Rightarrow \dfrac{{{n_R}}}{{{n_V}}} = \dfrac{{{\lambda _R}}}{{{\lambda _V}}}$
And we know that,
$ \Rightarrow {\lambda _R} > {\lambda _V}$
Then ${n_R}$is less than ${n_V}$
$\therefore {n_R} > {n_V}$
Hence the correct option is B.
Note: As we know that the different color of light has different wavelengths. In this question we use the concept that the wavelength of the red color light from the bulb is greater than the wavelength of the violet color light due to which the number of photons from the red color bulb will be greater than the number of photons of the violet color bulb.
Complete solution:
In this question, It is given that ${n_R}$ and ${n_V}$ denote the number of photons emitted by a red bulb and violet bulb of equal power in a given time. We need to find the relation between the number of photons emitted by a red bulb and by the violet bulb.
We will use the Planck law to get the energy as,
$ \Rightarrow E = \dfrac{{hc}}{\lambda }$
Here, the planck's constant is $h$, the speed of light is $c$, and the wavelength of the light is $\lambda $.
As we know that photons are the light whose move in packets and bundles.
If the power of each photon is $P$, then energy given out in t second is equal to $pt$. Let the number of photons be $n$, then
$ \Rightarrow n = \dfrac{{Pt}}{E}$
Now we will substitute the values in the above relation.
\[ \Rightarrow n = \dfrac{{Pt}}{{\left( {hc/\lambda } \right)}}\]
Now we simplify the above expression.
\[ \Rightarrow n = \dfrac{{Pt\lambda }}{{hc}}\]
For red light,
$ \Rightarrow {n_R} = \dfrac{{Pt{\lambda _R}}}{{hc}}$
For violet light,
$ \Rightarrow {n_V} = \dfrac{{Pt{\lambda _V}}}{{hc}}$
Since, from above relation we get,
$ \Rightarrow \dfrac{{{n_R}}}{{{n_V}}} = \dfrac{{{\lambda _R}}}{{{\lambda _V}}}$
And we know that,
$ \Rightarrow {\lambda _R} > {\lambda _V}$
Then ${n_R}$is less than ${n_V}$
$\therefore {n_R} > {n_V}$
Hence the correct option is B.
Note: As we know that the different color of light has different wavelengths. In this question we use the concept that the wavelength of the red color light from the bulb is greater than the wavelength of the violet color light due to which the number of photons from the red color bulb will be greater than the number of photons of the violet color bulb.
Recently Updated Pages
A square frame of side 10 cm and a long straight wire class 12 physics JEE_Main

The work done in slowly moving an electron of charge class 12 physics JEE_Main

Two identical charged spheres suspended from a common class 12 physics JEE_Main

According to Bohrs theory the timeaveraged magnetic class 12 physics JEE_Main

ill in the blanks Pure tungsten has A Low resistivity class 12 physics JEE_Main

The value of the resistor RS needed in the DC voltage class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

