A television tube operates at 20,000V. What is min wavelength for the continuous x-ray spectrum produced when the electrons hit the phosphor?
To find the minimum wavelength for the continuous X-ray spectrum produced when electrons hit the phosphor in a television tube operating at 20,000V, we can make use of the equations of X-ray wavelength and energy.
The energy of an X-ray photon can be calculated using the equation:
E = h * f
Where:
E = energy of the photon
h = Planck's constant (6.626 x 10^-34 J·s)
f = frequency of the X-ray
The frequency of the X-ray is related to the speed of light (c) and the wavelength (λ) through the equation:
f = c / λ
Rearranging the equation, we can express the wavelength as:
λ = c / f
Now, let's substitute the energy equation into the frequency equation to obtain:
λ = c / (E / h)
Considering that the voltage (V) of the television tube is related to the energy (E) using the equation:
E = V * e
Where:
V = voltage (20,000V in this case)
e = elementary charge (1.602 x 10^-19 C)
We can then replace E in the wavelength equation:
λ = c / (V * e / h)
The speed of light (c) is approximately 3 x 10^8 m/s. Substituting the known values:
λ = (3 x 10^8 m/s) / ((20,000V * 1.602 x 10^-19 C) / (6.626 x 10^-34 J·s))
Simplifying the equation:
λ = (3 x 10^8 m/s) * (6.626 x 10^-34 J·s) / (20,000V * 1.602 x 10^-19 C)
λ ≈ 3.89 x 10^-12 meters
Therefore, the minimum wavelength for the continuous X-ray spectrum produced when electrons hit the phosphor in the television tube is approximately 3.89 x 10^-12 meters.