Two sources of light emit with a power of 200 W.The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be :
- 1 : 5
- 1 : 3
- 5 : 3
- 3 : 5
The Correct Option is D
Approach Solution - 1
To determine the ratio of the number of photons emitted by each source, we need to first understand how the energy of light is related to the number of photons and their wavelength.
The energy \(E\) of a single photon is given by the equation:
\(E = \frac{hc}{\lambda}\)
where:
- \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \text{ J s}\)
- \(c\) is the speed of light, approximately \(3 \times 10^{8} \text{ m/s}\)
- \(\lambda\) is the wavelength of the light
The power \(P\) of the light source is related to the number of photons emitted per second \(N\) by:
\(P = N \times E = N \times \frac{hc}{\lambda}\)
Since both sources emit light with a power of 200 W, the equation for each source is:
\(200 = N_{1} \times \frac{hc}{300 \times 10^{-9}}\)
and
\(200 = N_{2} \times \frac{hc}{500 \times 10^{-9}}\)
We are tasked with finding the ratio \(\frac{N_{1}}{N_{2}}\):
- Solve the first equation for \(N_{1}\):
\(N_{1} = \frac{200 \cdot 300 \times 10^{-9}}{hc}\)
- Solve the second equation for \(N_{2}\):
\(N_{2} = \frac{200 \cdot 500 \times 10^{-9}}{hc}\)
- Calculate the ratio \(\frac{N_{1}}{N_{2}\):
\(\frac{N_{1}}{N_{2}} = \left(\frac{200 \cdot 300 \times 10^{-9}}{hc}\right) \div \left(\frac{200 \cdot 500 \times 10^{-9}}{hc}\right)\)
\(\frac{N_{1}}{N_{2}} = \frac{300}{500}\)
\(\frac{N_{1}}{N_{2}} = \frac{3}{5}\)
Thus, the ratio of the number of photons emitted by the two light sources is 3:5. Therefore, the correct answer is 3:5.
Approach Solution -2
Let \( n_1 \) and \( n_2 \) be the number of photons emitted by the sources with wavelengths \( \lambda_1 = 300 \, \text{nm} \) and \( \lambda_2 = 500 \, \text{nm} \), respectively.
Step 1. Calculate the energy of photons for each wavelength: The power emitted by each source is given as 200 W, so:
\(n_1 \times \frac{hc}{\lambda_1} = 200\)
\(n_2 \times \frac{hc}{\lambda_2} = 200\)
Step 2. Formulate the ratio:
\(\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}\)
Substituting values:
\(\frac{n_1}{n_2} = \frac{300}{500}\)
Simplifying, we get:
\(\frac{n_1}{n_2} = \frac{3}{5}\)
Therefore, the ratio of the number of photons emitted by each source is 3 : 5.
The Correct Answer is: 3:5
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Modern Physics Questions
- A long straight wire of circular cross-section (radius \( a \)) is carrying a steady current \( I \). The current \( I \) is uniformly distributed across this cross-section. The magnetic field is:
- A hydrogen atom is bombarded with electrons accelerated through a potential difference of \( V \), which causes excitation of hydrogen atoms. If the experiment is being performed at \( T = 0 \, \text{K} \), the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be\[\frac{\alpha}{10} V,\]where \( \alpha = \, \underline{\hspace{2cm}} \).
- The energy equivalent of 1g of substance is:
- From the statements given below :
(A) The angular momentum of an electron in nth orbit is an integral multiple of h.
(B) Nuclear forces do not obey inverse square law.
(C) Nuclear forces are spin dependent.
(D) Nuclear forces are central and charge independent.
(E) Stability of nucleus is inversely proportional to the value of packing fraction.
Choose the correct answer from the options given below : - Work function for an object is 2.3 eV. If maximum kinetic energy of ejected electrons is 0.18 eV, find wavelength \( \lambda \) of incident photon on object.
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.