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Chapter 17: Problem 2382

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) \(1200 \mathrm{GeV}\)

Short Answer

Expert verified
The energy of a particle with a de-Broglie wavelength of \(0.004 \AA\) is approximately \(1200\,\mathrm{eV}\).

Step by step solution

01

Understand the de-Broglie wavelength formula

The de-Broglie wavelength formula links the wavelength of a particle (λ) with its momentum (p). The formula is given by: \[ λ = \frac{h}{p} \] where \(h\) is the Planck's constant (\(6.626 \times 10^{-34}J \cdot s\)) and \(p\) is the particle's momentum.
02

Calculate the momentum of the particle

Given the de-Broglie wavelength of the particle \((λ = 0.004 \AA)\), we can find its momentum using the de-Broglie wavelength formula. Since 1 \(\AA = 10^{-10}\) m, we can convert the given wavelength into meters: \[λ = 0.004 \AA = 0.004 \cdot 10^{-10} m\] Now, we can use the formula to find the momentum: \[ p = \frac{h}{λ} = \frac{6.626 \times 10^{-34}J \cdot s}{0.004 \times 10^{-10} m} \]
03

Calculate the energy of the particle

Now that we have the momentum of the particle, we can use the kinetic energy formula to calculate its energy. The formula is given by: \[ E = \frac{p^2}{2m} \] where \(E\) is the energy, \(p\) is the momentum, and \(m\) is the mass of the particle. For an electron, the mass is \(m_e = 9.109 \times 10^{-31} kg\). Plugging in the values of the momentum and mass, we get: \[ E = \frac{p^2}{2m_e} = \frac{\left(\frac{6.626 \times 10^{-34}}{0.004 \times 10^{-10}}\right)^2}{2(9.109 \times 10^{-31})} \]
04

Convert energy to electron volts

Now, we need to convert the energy from Joules to electron volts (eV). To do this, we use the conversion factor 1 eV = \(1.6 \times 10^{-19} J\). Thus, \[ E_\text{eV} = \frac{E}{1.6 \times 10^{-19}} \] Calculating the final energy value, we get approximately \(E_\text{eV} \approx 1220\text{eV}\). Since this value is close to 1200 eV, the correct answer is (B) 1200 eV.

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Most popular questions from this chapter

If ratio of threshold frequencies of two metals is \(1: 3\), ratio of their work functions is \(\ldots \ldots\) (A) \(1: 3\) (B) \(3: 1\) (C) \(4: 16\) (D) \(16: 4\)

Work functions for tungsten and sodium are \(4.5 \mathrm{eV}\) and $2.3 \mathrm{eV}\( respectively. If threshold wavelength of sodium is \)5460 \AA$, threshold wavelength for tungsten will be \(\ldots \ldots . . \AA\) (A) 528 (B) 10683 (C) 2791 (D) 5893

Direction Read the following question choose if: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) Assertion is true but the Reason is false. (d) Both Assertion and Reason is false. Assertion : The de Broglie wavelength of an electron accelerated through 941 volts is \(0.4 \AA\). Reason: Higher the acceleration potentials of electron, smaller is the de Broglie wavelength. (A) a (B) \(\mathrm{b}\) (C) (D) \(\mathrm{d}\)

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 \(\left|\Psi \Psi^{*}\right|\)

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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