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An x-ray tube has an applied voltage of 640kV.

(a) What is the most energetic x-ray photon it can produce? Express your answer in electron volts.

How does energy relate to the voltage? eV

(b) Find the wavelength (in nm ) of such an x-ray.

How does energy relate to the wavelength? nm

User Gabr
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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).

User Aswin Kumar
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