Question

Alpha particles are accelerated through a potential difference of \(20.0 \mathrm{kV}\). What is their de Broglie wavelength?

Short answer

Question: Calculate the de Broglie wavelength of alpha particles accelerated through a potential difference of 20.0 kV.

Step-by-step solution

Find the kinetic energy of the alpha particles

To find the kinetic energy (KE) of the alpha particles, we can use the equation: KE = q * V Where q is the charge of the alpha particles and V is the potential difference through which they are accelerated. Alpha particles have a charge of +2e, where e is the elementary charge (approximately \(1.602 \times 10^{-19} \mathrm{C}\)). The potential difference is given as 20.0 kV or \(20.0 \times 10^3 \mathrm{V}\). KE = (2e) * (20.0 kV)= \(20.0 \times 10^3 \mathrm{V} \times 2 \times 1.602 \times 10^{-19} \mathrm{C}\)

Calculate the velocity of the alpha particles

Now that we have the kinetic energy, we can find the velocity of the alpha particles using the equation: KE = \(\frac{1}{2} m v^2\) Where m is the mass of an alpha particle (approximately \(6.644 \times 10^{-27} \mathrm{kg}\)) and v is the velocity of the alpha particles. Rearranging the equation to find the velocity, we get: v = \(\sqrt{\frac{2 \times KE}{m}}\) Plugging in the values, we get: v = \(\sqrt{\frac{2 \times (20.0 \times 10^3 \mathrm{V} \times 2 \times 1.602 \times 10^{-19} \mathrm{C})}{6.644 \times 10^{-27} \mathrm{kg}}}\)

Calculate the de Broglie wavelength

With the velocity of the alpha particles, we can now calculate their de Broglie wavelength using the equation: \(\lambda = \frac{h}{m v}\) Where \(\lambda\) is the de Broglie wavelength, h is the Planck's constant (approximately \(6.626 \times 10^{-34} \mathrm{Js}\)), and v is the velocity of the alpha particles. Applying the velocity calculated in step 2: \(\lambda = \frac{6.626 \times 10^{-34} \mathrm{Js}}{(6.644 \times 10^{-27} \mathrm{kg})\sqrt{\frac{2 \times (20.0 \times 10^3 \mathrm{V} \times 2 \times 1.602 \times 10^{-19} \mathrm{C})}{6.644 \times 10^{-27} \mathrm{kg}}}}\) Calculating these values gives the de Broglie wavelength of the alpha particles.