Chapter 8: Problem 77

A color television tube generates some X-rays when its electron beam strikes the screen. What is the shortest wavelength of these X-rays, if a 30.0 -kV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these X-rays from exposing viewers.)

Short Answer

Expert verified

The shortest wavelength of the X-rays generated when the electron beam strikes the screen in a color television tube with a 30.0 kV potential is \(4.13 × 10^{-11} m\).

Step by step solution

Apply energy conservation principle and the equation of a photon's energy

Since the electrons are accelerated by the 30.0 kV potential, the energy gained by them will be converted into the energy of the X-rays. Using energy conservation, we can write the equation: \(E_{electric} = E_{x-ray}\) The energy of a photon, \(E_{x-ray}\), is given by the following equation: \(E_{x-ray} = h \cdot f\) where h is the Planck's constant (6.63 × 10^(-34) J·s) and f is the frequency of the X-ray. The frequency can be expressed in terms of the wavelength λ as: \(f = \frac{c}{λ}\) where c is the speed of light (3.00 × 10^(8) m/s).

Calculate the energy gained by the electron

The energy gained by the electron can be calculated using the following equation: \(E_{electric} = e \cdot V\) where e is the charge of an electron (1.60 × 10^(-19) C) and V is the potential difference (30.0 kV = 30,000 V). Now, substitute the given values into the equation and solve for the energy gained by the electron: \(E_{electric} = (1.60 × 10^{-19} C) \cdot (30,000 V)\) \(E_{electric} = 4.80 × 10^{-15} J\)

Relate the energy gained by the electron to the energy of the X-ray and find the wavelength

Now we know that the energy gained by the electrons should be equal to the energy of the X-ray: \(E_{electric} = E_{x-ray}\) From step 1 and 2, we have: \(4.80 × 10^{-15} J = h \cdot \frac{c}{λ}\) Now, solve for the wavelength λ by substituting the values for h and c: \(λ = \frac{h \cdot c}{E_{electric}}\) \(λ = \frac{(6.63 × 10^{-34} J·s) \cdot (3.00 × 10^{8} m/s)}{4.80 × 10^{-15} J}\) \(λ = 4.13 × 10^{-11} m\) So, the shortest wavelength of these X-rays is \(4.13 × 10^{-11} m\).

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A color television tube generates some X-rays when its electron beam strikes the screen. What is the shortest wavelength of these X-rays, if a 30.0 -kV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these X-rays from exposing viewers.)

Short Answer

Step by step solution

Apply energy conservation principle and the equation of a photon's energy

Calculate the energy gained by the electron

Relate the energy gained by the electron to the energy of the X-ray and find the wavelength

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