Question

In a color TV tube, electrons are accelerated through a potential difference of \(20.0 \mathrm{kV} .\) Some of the electrons strike the metal mask (instead of the phosphor dots behind holes in the mask), causing x-rays to be emitted. What is the smallest wavelength of the \(x\) -rays emitted?

Short answer

Answer: The smallest wavelength of the x-rays emitted is 0.062 nm.

Step-by-step solution

Identify the Given Information and Formula to be Used

We are given: - Potential Difference, V = 20.0 kV or 20000 V We need to find the smallest wavelength of the x-rays emitted. We will use the following formulas: 1. Energy of accelerated electrons: E = eV (where e is the electron charge and V is the potential difference) 2. Planck's formula: E = h * c / λ (where h is the Planck's constant, c is the speed of light, and λ is the wavelength) 3. Bragg's condition- Wavelength-energy relation for the smallest wavelength.

Find the Energy of Accelerated Electrons

Firstly, we need to find the energy of the electrons when they are accelerated with a potential difference of 20.0 kV. We can use the formula: E = eV Here, e is the electron charge, which is approximately \(1.6 * 10^{-19} \,\text{C}\). The potential difference, V, is given as 20.0 kV or 20000 V. E = e * V = \((1.6 * 10^{-19} \,\text{C}) * (20000\,\text{V})\)= \(3.2 * 10^{-15}\,\text{J}\) So, the energy of the accelerated electrons is \(3.2 * 10^{-15}\,\text{J}\).

Using Planck's Formula to find the Wavelength

Next, we will use the Planck's Formula to find the wavelength: E = h * c / λ Here, E = \(3.2 * 10^{-15}\,\text{J}\) (the energy of the electrons) h = \(6.626 * 10^{-34}\,\text{J}\cdot\text{s}\) (Planck's constant) c = \(3 * 10^8\,\text{m}/\text{s}\) (Speed of light) Rearrange the formula to find the wavelength: λ = h * c / E Plug in the values: λ = \((6.626 * 10^{-34}\,\text{J}\cdot\text{s}) * (3 * 10^8\,\text{m}/\text{s}) / (3.2 * 10^{-15}\,\text{J})\) λ = \(6.206 * 10^{-11}\,\text{m}\)

Convert the Wavelength to Nanometers and find the Smallest Wavelength

Finally, we will convert the wavelength from meters to nanometers (1 m = \(10^9\) nm) and find the smallest possible wavelength: The smallest wavelength (in nm): λ_min = \(6.206 * 10^{-11}\,\text{m} * 10^9\,\text{nm}/\text{m}\) = \(0.062\,\text{nm}\) Therefore, the smallest wavelength of the x-rays emitted is \(0.062\,\text{nm}\).