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Chapter 17: Problem 2322

A body of mass \(200 \mathrm{~g}\) moves at the speed of $5 \mathrm{~m} / \mathrm{hr}$. So deBroglie wavelength related to it is of the order........ \(\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\) (A) \(10^{-10} \mathrm{~m}\) (B) \(10^{-20} \mathrm{~m}\) (C) \(10^{-30} \mathrm{~m}\) (D) \(10^{-40} \mathrm{~m}\)

Short Answer

Expert verified
The de Broglie wavelength related to the given body is of the order \(10^{-30} \mathrm{~m}\).

Step by step solution

01

Convert mass to kilograms

The mass is given as 200 g, which can be converted to kg by dividing by 1000: Mass = \(200 \mathrm{~g} \div 1000 = 0.2 \mathrm{~kg}\)
02

Convert speed to meters per second

The speed is given as 5 m/hr, which can be converted to m/s by multiplying by the conversion factor \(1/3600\): Speed = \(5 \mathrm{~m/hr} \times \frac{1}{3600} = \frac{5}{3600} \mathrm{~m/s}\) #Step 2: Use the de Broglie wavelength formula#
03

Recall the de Broglie formula

The de Broglie wavelength formula is given by: Wavelength = \(\frac{h}{p}\) Where 'h' is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\)), 'p' is the momentum of the body which is given by the product of its mass and speed. #Step 3: Calculate the momentum of the body#
04

Calculate momentum

Momentum is the product of mass and speed: Momentum (p) = Mass × Speed = \(0.2 \mathrm{~kg} \times \frac{5}{3600} \mathrm{~m/s} = \frac{1}{3600} \mathrm{~kg} \cdot \mathrm{m/s}\) #Step 4: Calculate the de Broglie wavelength#
05

Calculate the de Broglie wavelength

Using the de Broglie formula and the momentum calculated in step 3: Wavelength = \(\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\frac{1}{3600} \mathrm{~kg} \cdot \mathrm{m/s}} = 2.403 \times 10^{-31} \mathrm{~m}\) Since the given options are in the order of magnitude, we can find out that the closest value is in the order of \(10^{-30} \mathrm{~m}\). Answer: (C) \(10^{-30} \mathrm{~m}\)

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Most popular questions from this chapter

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) \(8.4\)

Work function of metal is \(2.5 \mathrm{eV}\) If wave length of light incident on metal plate is \(3000 \AA\), stopping potential of emitted electron will be....... $\left\\{\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(0.82 \mathrm{~V}\) (B) \(0.41 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(3.28 \mathrm{~V}\)

Output power of He-Ne LASER of low energy is \(1.00 \mathrm{~mW}\). Wavelength of the light is \(632.8 \mathrm{~nm}\). What will be the number of photons emitted per second from this LASER? (A) \(8.31 \times 10^{15} \mathrm{~s}^{-1}\) (B) \(5.38 \times 10^{15} \mathrm{~s}^{-1}\) (C) \(1.83 \times 10^{15} \mathrm{~s}^{-1}\) (D) \(3.18 \times 10^{15} \mathrm{~s}^{-1}\)

Compare energy of a photon of X-rays having \(1 \AA\) wavelength with the energy of an electron having same de Broglie wavelength. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, 1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(8.24\) (B) \(2.48\) (C) \(82.4\) (D) \(24.8\)

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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