Answer: (a) To find the most energetic X-ray photon that the X-ray tube can produce, we can use the relationship between photon energy (E) and voltage (V) in electron volts (eV):
Energy (E) of a photon = qV
where q is the elementary charge (approximately 1.602 x 10^-19 coulombs) and V is the voltage.
Given that the applied voltage (V) is 640 kV (640,000 volts), we can calculate the most energetic X-ray photon's energy as follows:
Energy (E) = q * V
E = 1.602 x 10^-19 C * 640,000 V
E ≈ 1.02528 x 10^-13 joules
Now, let's convert the energy from joules to electron volts (eV):
1 eV = 1.602 x 10^-19 joules
Energy (E) in eV ≈ (1.02528 x 10^-13 joules) / (1.602 x 10^-19 joules/eV) ≈ 639.98 eV
So, the most energetic X-ray photon produced by the X-ray tube has an energy of approximately 639.98 electron volts (eV).
(b) To find the wavelength (in nm) of such an X-ray, we can use the relationship between photon energy (E) and wavelength (λ) for electromagnetic waves, including X-rays:
Energy (E) of a photon = h * c / λ
where h is Planck's constant (approximately 6.626 x 10^-34 joule-seconds) and c is the speed of light in a vacuum (approximately 3.00 x 10^8 meters per second).
We already know the energy of the photon (E ≈ 1.02528 x 10^-13 joules). Now, let's find the wavelength (λ) in meters and then convert it to nanometers:
1 meter = 1 x 10^9 nanometers (nm)
λ = h * c / E
λ ≈ (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (1.02528 x 10^-13 J)
λ ≈ 1.54 x 10^-11 meters
Now, convert the wavelength to nanometers:
λ ≈ 1.54 x 10^-11 meters * (1 x 10^9 nm/meter) ≈ 15.4 nm
The wavelength of the X-ray produced by the X-ray tube is approximately 15.4 nanometers (nm).