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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

Energy of photon of light having two different frequencies are \(2 \mathrm{eV}\) and \(10 \mathrm{eV}\) respectively. If both are incident on the metal having work function \(1 \mathrm{eV}\), ratio of maximum velocities of emitted electron is ............ (A) \(1: 5\) (B) \(3: 11\) (C) \(2: 9\) (D) \(1: 3\)

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

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) \(4395 \AA\)

Uncertainty in position of electron is found of the order of de-Broglie wavelength. Using Heisenberg's uncertainty principle, it is found that order of uncertainty in its velocity \(=\ldots \ldots \ldots .\) (A) \(1 \mathrm{v}\) (B) \(2 \mathrm{v}\) (C) \((\mathrm{v} / 2 \pi)\) (D) \(2 \pi \mathrm{v}\)

Radius of a nucleus \(2 \times 10^{-15} \mathrm{~m} .\) If we imagine an electron inside the nucleus then energy of electron will be $=\ldots \ldots . \mathrm{MeV}$ $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ (A) \(6.98 \times 10^{3}\) (B) \(8.94 \times 10^{3}\) (C) \(4.98 \times 10^{3}\) (D) \(9.48 \times 10^{3}\)
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