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Chapter 17: Problem 2439

If \(\propto\) -particle and proton have same velocities, the ratio of de Broglie wavelength of \(\propto\) -particle and proton is \(\ldots \ldots\) (A) \((1 / 4)\) (B) \((1 / 2)\) (C) 1 (D) 2

Short Answer

Expert verified
The ratio of de Broglie wavelength of 𝛼-particle and proton when they have the same velocities is \(\frac{1}{4}\).

Step by step solution

01

Find the de Broglie wavelength formula

The de Broglie wavelength formula states that the wavelength (λ) is related to the momentum (p) of a particle using Planck’s constant (h) as follows: \[λ = \frac{h}{p}\]
02

Express momentum in terms of mass and velocity

We know that the momentum (p) can be expressed as the product of the mass (m) and velocity (v) of a particle: \[p = m \cdot v\]
03

Substitute momentum expression into de Broglie wavelength formula

Using our expression for momentum, we can rewrite the de Broglie wavelength formula as follows: \[λ = \frac{h}{m \cdot v}\]
04

Set up the ratio for the wavelengths of 𝛼-particle and proton

Let λ_α and λ_p be the de Broglie wavelengths of the 𝛼-particle and proton, respectively. We can set up a ratio as follows: \[\frac{λ_α}{λ_p} = \frac{\frac{h}{m_α \cdot v}}{\frac{h}{m_p \cdot v}}\]
05

Simplify the ratio

Since the 𝛼-particle and proton have the same velocity, we can simplify the ratio by canceling out the "h" and "v" terms and solving for the mass ratio: \[\frac{λ_α}{λ_p} = \frac{m_p}{m_α} \]
06

Use given masses for 𝛼-particle and proton to find the ratio

Given that the mass of 𝛼-particle (m_α) is 4 times the mass of a proton (m_p), we can substitute this information into the ratio formula: \[\frac{λ_α}{λ_p} = \frac{m_p}{4 \cdot m_p}\]
07

Calculate the final ratio

By canceling out m_p, we are left with the final ratio of the de Broglie wavelengths of 𝛼-particle and proton: \[\frac{λ_α}{λ_p} = \frac{1}{4}\] Thus, the correct answer is: (A) \((\frac{1}{4})\)

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Most popular questions from this chapter

Suppose \(\Psi(\mathrm{x}, \mathrm{y}, \mathrm{z})\) represents a particle in three dimensional space, then probability of finding the particle in the unit volume at a given point \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) is $\ldots \ldots$ (A) inversely proportional to $\Psi^{\prime}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ (B) directly proportional \(\Psi^{*}\) (C) directly proportional to \(\mid \Psi \Psi^{*}\) (D) inversely proportional to \(\left|\Psi \Psi^{*}\right|\)

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Read the paragraph carefully and select the proper choice from given multiple choices. According to Einstein when a photon of light of frequency for wavelength \(\lambda\) is incident on a photo sensitive metal surface of work function \(\Phi\). Where \(\Phi<\mathrm{hf}\) (here \(\mathrm{h}\) is Plank's constant) then the emission of photo-electrons place takes place. The maximum K.E. of emitted photo electrons is given by $\mathrm{K}_{\max }=\mathrm{hf}-\Phi .\( If the there hold frequency of metal is \)\mathrm{f}_{0}$ then \(\mathrm{hf}_{0}=\Phi\) (i) A metal of work function \(3.3 \mathrm{eV}\) is illuminated by light of wave length \(300 \mathrm{~nm}\). The maximum \(\mathrm{K} . \mathrm{E}>\) of photo- electrons is $=\ldots \ldots \ldots . \mathrm{eV} .\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}\right)$ (A) \(0.825\) (B) \(0.413\) (C) \(1.32\) (D) \(1.65\) (ii) Stopping potential of emitted photo-electron is $=\ldots \ldots . \mathrm{V}$. (A) \(0.413\) (B) \(0.825\) (C) \(1.32\) (D) \(1.65\) (iii) The threshold frequency fo $=\ldots \ldots \ldots \times 10^{14} \mathrm{~Hz}$. (A) \(4.0\) (B) \(4.2\) (C) \(8.0\) (D) \(8.4\)

The de-Broglie wavelength of a proton and \(\alpha\) - particle is same. The ratio of their velocities will be.......... ( \(\alpha\) particle is the He- nucleus, having two protons and two neutrons. Thus, its mass \(\mathrm{M}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\) where \(\mathrm{m}_{\mathrm{p}}\) is the mass of the proton.) (A) \(1: 4\) (B) \(1: 2\) (C) \(2: 1\) (D) \(4: 1\)
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