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Chapter 17: Problem 2381

An electron is accelerated between two points having potential $20 \mathrm{~V}\( and \)40 \mathrm{~V}$, de-Broglic wavelength of electron is \(\ldots \ldots\) (A) \(0.75 \AA\) (B) \(7.5 \AA\) (C) \(2.75 \AA\) (D) \(0.75 \mathrm{~nm}\)

Short Answer

Expert verified
The de Broglie wavelength of the electron is approximately \(2.75 \AA\).

Step by step solution

01

Calculate the potential difference (V) between the two points.

We are given two potentials: 20V and 40V. Subtract the lower potential from the higher potential to calculate the potential difference: V = 40V - 20V = 20V
02

Find the kinetic energy gained by the electron.

The electron gains kinetic energy as it moves through the potential difference. We can use the formula: K.E. = eV Where e is the electron charge (1.6 x 10^{-19} C) and V is the potential difference. Plug in the values to find the kinetic energy: K.E. = (1.6 x 10^{-19} C) × (20 V) = 3.2 x 10^{-18} J
03

Use the kinetic energy to find the electron's momentum.

Since the electron gains kinetic energy, we can use the relation between kinetic energy and momentum: K.E. = (1/2)mv^2 Where m is the mass of the electron (9.11 x 10^{-31} kg) and v is its velocity. However, we are interested in the momentum (p) of the electron, which is given by p=mv. We can rewrite the kinetic energy equation in terms of momentum: K.E. = (p^2)/(2m) Now, we can solve for the momentum: p^2 = 2m × K.E. p = \(\sqrt{2m \times K.E.}\) Plug in the values to find the momentum: p = \(\sqrt{2(9.11 \times 10^{-31} kg)(3.2 \times 10^{-18} J)}\) = 1.96 × 10^{-24} kg·m/s
04

Calculate the de Broglie wavelength of the electron.

Now that we have the momentum of the electron, we can use the de Broglie wavelength formula to find its wavelength: λ = h / p Where λ is the de Broglie wavelength, h is the Planck constant (6.626 x 10^{-34} Js), and p is the electron's momentum. Plug in the values to find the wavelength: λ = (6.626 × 10^{-34} Js) / (1.96 × 10^{-24} kg·m/s) = 3.38 × 10^{-10} m
05

Convert the wavelength to the appropriate unit and compare with the options.

The calculated de Broglie wavelength λ is in meters, but the given options are in Ångström (Å) or nanometers (nm). We need to convert the wavelength to the appropriate unit for comparison. 1Å = 10^{-10} m, and 1nm = 10^{-9} m. λ = 3.38 × 10^{-10} m = 3.38 Å (convert to Å) Compare the result with the given options: (A) 0.75 Å (B) 7.5 Å (C) 2.75 Å (D) 0.75 nm Our calculated de Broglie wavelength is closest to option (C), 2.75 Å. Therefore, the answer is (C) \(2.75 \AA\).

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Most popular questions from this chapter

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