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Chapter 17: Problem 2369

Ration of momentum of photons having wavelength \(4000 \AA \& 8000 \AA\) is ........... (A) \(2: 1\) (B) \(1: 2\) (C) \(20: 1\) (D) \(1: 20\)

Short Answer

Expert verified
The ratio of momenta of photons having wavelengths 4000 Å and 8000 Å is 2:1. The correct answer is (A) \(2: 1\).

Step by step solution

01

Convert Wavelengths

First, let's convert the given wavelengths from Angstrom to meters. 1 Å = \(10^{-10}\) meters So the wavelengths in meters are: λ₁ = 4000 Å = 4000 × \(10^{-10}\) m = \(4 \times 10^{-7}\) m λ₂ = 8000 Å = 8000 × \(10^{-10}\) m = \(8 \times 10^{-7}\) m
02

Calculate Momentum

Now, let's find the momentum of the two photons using the formula: momentum (p) = \(\frac{h}{λ}\) where h is the Planck constant, h = \(6.63 \times 10^{-34}\) Js For Photon 1 with λ₁: \(p_1 = \frac{6.63 \times 10^{-34}}{4 \times 10^{-7}}\) For Photon 2 with λ₂: \(p_2 = \frac{6.63 \times 10^{-34}}{8 \times 10^{-7}}\)
03

Find the Ratio

Now, we will find the ratio of the momenta, \(\frac{p_1}{p_2} = \frac{\frac{6.63 \times 10^{-34}}{4 \times 10^{-7}}}{\frac{6.63 \times 10^{-34}}{8 \times 10^{-7}}}\) Notice that the Planck constant (h) on both numerator and denominator cancels out: \(\frac{p_1}{p_2} = \frac{\frac{1}{4 \times 10^{-7}}}{\frac{1}{8 \times 10^{-7}}}\) Now, we can simplify the ratio by multiplying both numerator and denominator by \(8 \times 10^{-7}\): \(\frac{p_1}{p_2} = \frac{2}{1}\)
04

Final Answer

Thus, the ratio of momenta of photons having wavelengths 4000 Å and 8000 Å is 2:1. The correct answer is (A) \(2: 1\).

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The de-Broglie wavelength associated with a particle with rest mass \(\mathrm{m}_{0}\) and moving with speed of light in vacuum is..... (A) \(\left(\mathrm{h} / \mathrm{m}_{0} \mathrm{c}\right)\) (B) 0 (C) \(\infty\) (D) \(\left(\mathrm{m}_{0} \mathrm{c} / \mathrm{h}\right)\)

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