Question

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

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

Step-by-step solution

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

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

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

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

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) \(2.75 \AA\).