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Chapter 17: Problem 2327

If kinetic energy of free electron is made double, change in de-Broglie wavelength will be........... (A) \(\sqrt{2}\) (B) \((1 / \sqrt{2})\) (C) 2 (D) \((1 / 2)\)

Short Answer

Expert verified
The change in de-Broglie wavelength will be \( \frac{1}{\sqrt{2}}\) (B).

Step by step solution

01

Find the initial wavelength

First, let's use the initial kinetic energy to determine the initial de-Broglie wavelength. Using the relationship between kinetic energy and velocity, we can write the equation: $$ \frac{1}{2}mv^2 = K_{1} $$ where \(K_{1}\) is the initial kinetic energy.
02

Find the final wavelength

Now, let's use the final kinetic energy (which is double the initial kinetic energy) to determine the final de-Broglie wavelength. We can write the equation: $$ \frac{1}{2}mv'^2 = 2K_{1} $$ where \(v'\) is the final velocity.
03

Find the change in the de-Broglie wavelength

Now, we need to find the ratio \(\frac{\lambda'}{\lambda}\), where \(\lambda'\) is the final de-Broglie wavelength and \(\lambda\) is the initial de-Broglie wavelength. Using the equations from steps 1 and 2, we have: $$ \frac{\lambda'}{\lambda} = \frac{\frac{h}{mv'}}{\frac{h}{mv}} = \frac{v}{v'} $$
04

Solve the equation for the change in wavelength

We can now solve for the velocity ratio using the kinetic energy equations from steps 1 and 2: $$ v'^2 = 4v^2 $$ $$ \frac{v'}{v} = \frac{1}{\sqrt{2}} $$ Therefore, the change in de-Broglie wavelength will be: $$ \frac{\lambda'}{\lambda} = \boxed{\frac{1}{\sqrt{2}}} \text{ (B)} $$

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Most popular questions from this chapter

Energy of a particle having de-Broglie wavelength \(0.004 \AA\) is... (A) \(1280 \mathrm{eV}\) (B) \(1200 \mathrm{eV}\) (C) \(1200 \mathrm{MeV}\) (D) \(1200 \mathrm{GeV}\)

Radius of a beam of radiation of wavelength \(5000 \AA\) is $10^{-3} \mathrm{~m}\(. Power of the beam is \)10^{-3} \mathrm{~W}$. This beam is normally incident on a metal of work function \(1.9 \mathrm{eV}\). The charge emitted by the metal per unit area in unit time is \(\ldots \ldots \ldots\) Assume that each incident photon emits one electron. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(1.282 \mathrm{C}\) (B) \(12.82 \mathrm{C}\) (C) \(128.2 \mathrm{C}\) (D) \(1282 \mathrm{C}\)

When a radiation of wavelength \(3000 \AA\) is incident on metal, $1.85 \mathrm{~V}$ stopping potential is obtained. What will be threshed wave length of metal? $\left\\{\mathrm{h}=66 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(4539 \AA\) (B) \(3954 \AA\) (C) \(5439 \AA\) (D) \(4395 \AA\)

Frequency of photon having energy \(66 \mathrm{eV}\) is ....... \(\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)\) (A) \(8 \times 10^{-15} \mathrm{~Hz}\) (B) \(12 \times 10^{-15} \mathrm{~Hz}\) (C) \(16 \times 10^{-15} \mathrm{~Hz}\) (D) \(24 \times 10^{+15} \mathrm{~Hz}\)

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) \(8.4\)
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