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Chapter 17: Problem 2335

How many photons of red colored light having wavelength \(8000 \AA\) will have same energy as one photon of violet colored light of wavelength \(4000 \AA\) ? (A) 2 (B) 4 (C) 6 (D) 8

Short Answer

Expert verified
\(n = 2\), so 2 photons of red-colored light have the same energy as one photon of violet-colored light. The correct answer is (A).

Step by step solution

01

Planck's Equation for Photon Energy

The photon energy can be calculated using Planck's equation, which is given by: \[E= \dfrac{hc}{\lambda}\] Where \(E\) is the energy of the photon, \(h\) is the Planck's constant \((6.626 \times 10^{-34} Js)\), \(c\) is the speed of light \((3 \times 10^8 m/s)\), and \(\lambda\) is the wavelength of the photon.
02

Calculate the energy of violet and red photon

Using Planck's equation, we can first calculate the energy of one violet photon and then the energy of a red photon: \[E_v = \dfrac{hc}{\lambda_v}\] and \[E_r = \dfrac{hc}{\lambda_r}\] Here, the wavelengths are given in angstroms (Å). We need to convert them to meters before calculating the energy. The conversion factor is \(1 Å = 1 \times 10^{-10} m\).
03

Ratio of energies of red and violet photons

Divide the energy equation of the red photon by the energy equation of the violet photon to find the ratio: \[\dfrac{E_r}{E_v} = \dfrac{\dfrac{hc}{\lambda_r}}{\dfrac{hc}{\lambda_v}}\] Simplifying this expression, the Planck's constant and the speed of light cancel out, leaving: \[\dfrac{E_r}{E_v} = \dfrac{\lambda_v}{\lambda_r}\]
04

Calculate the number of red photons

Now, we want to find out how many red photons will have the same energy as one violet photon. Let's call this number \(n\). Then, substitute the respective wavelengths of the red and violet photons into this equation: \[n \cdot E_r = E_v\] Using the previous result, we get: \[n \cdot \dfrac{\lambda_v}{\lambda_r} = 1\] Now, plug in the wavelengths for red and violet light in meters: \[n \cdot \dfrac{4000 \times 10^{-10}}{8000 \times 10^{-10}} = 1\]
05

Solve for n and compare with options

Now solve for \(n\): \[n = \dfrac{8000 \times 10^{-10}}{4000 \times 10^{-10}}\] \[n = 2\] So, 2 photons of red-colored light have the same energy as one photon of violet-colored light. Hence, the correct answer is (A).

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Most popular questions from this chapter

Wavelength of an electron having energy \(10 \mathrm{ke} \mathrm{V}\) is $\ldots \ldots . \AA$ (A) \(0.12\) (B) \(1.2\) (C) 12 (D) 120

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In photoelectric effect, work function of martial is \(3.5 \mathrm{eV}\). By applying \(-1.2 \mathrm{~V}\) potential, photo electric current becomes zero, so......... (A) energy of incident photon is \(4.7 \mathrm{eV}\). (B) energy of incident photon is \(2.3 \mathrm{eV}\) (C) If photon having higher frequency is used, photo electric current is produced. (D) When energy of photon is \(2.3 \mathrm{eV}\), photo electric current becomes maximum

Through what potential difference should an electron be accelerated so it's de-Broglie wavelength is \(0.3 \AA\). (A) \(1812 \mathrm{~V}\) (B) \(167.2 \mathrm{~V}\) (C) \(1516 \mathrm{~V}\) (D) \(1672.8 \mathrm{~V}\)

Wavelength \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\) are incident on two identical metal plates and photo electrons are emitted. If \(\lambda_{\mathrm{A}}=2 \lambda_{\mathrm{B}}\), the maximum kinetic energy of photo electrons is \(\ldots \ldots \ldots\) (A) \(2 \mathrm{~K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}\) (B) \(\mathrm{K}_{\mathrm{A}}<\left(\mathrm{K}_{\mathrm{B}} / 2\right)\) (C) \(\mathrm{K}_{\mathrm{A}}=2 \mathrm{~K}_{\mathrm{B}}\) (D) \(\mathrm{K}_{\mathrm{A}}>\left(\mathrm{K}_{\mathrm{B}} / 2\right)\)
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