A photon of X ray has energy of 1 keV, while a photon of visible light has energy of about 3 eV. In this context, which one of the following statements is not correct?

Introduction / Context:
Photons are quanta of electromagnetic radiation, and their properties such as energy, wavelength, and frequency are related. X rays are high energy electromagnetic waves, while visible light has lower photon energy. Although their energies and wavelengths differ, all electromagnetic waves share certain fundamental properties when they travel in vacuum. This question asks you to identify which of the listed statements about an X ray photon and a visible light photon is not correct.

Given Data / Assumptions:

• Energy of the X ray photon E x = 1 keV (1000 eV).
• Energy of the visible light photon E v = 3 eV.
• Both photons travel in vacuum when we compare their speeds.
• Standard relations E = h * f and c = lambda * f apply, where h is Planck constant, f is frequency, lambda is wavelength, and c is speed of light in vacuum.

Concept / Approach:
Photon energy E is proportional to frequency f through the relation E = h * f. Therefore, a larger energy photon has higher frequency. Since the speed of light in vacuum c is constant for all electromagnetic waves, wavelength lambda is related by c = lambda * f, so higher frequency implies shorter wavelength. Thus, a 1 keV X ray photon has much higher frequency and shorter wavelength than a 3 eV visible photon. However, the speed in vacuum is the same c for both. Any statement claiming different speeds in vacuum is therefore incorrect.

Step-by-Step Solution:
Step 1: Compare energies: 1 keV equals 1000 eV, while the visible photon has 3 eV. So E x is much greater than E v. Step 2: Use E = h * f. Higher energy means higher frequency. Therefore, the X ray photon has higher frequency than the visible photon. Step 3: Use c = lambda * f. For a given constant c, a higher frequency implies a shorter wavelength. So the X ray photon has a shorter wavelength than the visible photon. Step 4: Recall that all electromagnetic waves in vacuum travel at the same speed c, regardless of their frequency or wavelength. This applies to X rays, visible light, radio waves, and gamma rays alike. Step 5: Therefore, any statement that claims that the speeds of the two photons in vacuum are different is not correct.

Verification / Alternative check:
Standard physics references define c, the speed of light in vacuum, as a universal constant approximately equal to 3 * 10^8 m/s. This value applies to all electromagnetic waves in vacuum. Differences in speed occur only when waves travel through media with different refractive indices, and even then the speed depends on the medium, not on photon energy alone. Since the question explicitly states speeds in vacuum, c must be the same. On the other hand, X ray wavelengths are well known to be much shorter (for example, nanometres to picometres) than visible wavelengths (hundreds of nanometres), consistent with the first statement being correct.

Why Other Options Are Wrong (meaning they are actually correct statements):
The wavelength of the X ray photon is less than the wavelength of the visible light photon: This is correct because higher energy and higher frequency correspond to shorter wavelength for the X ray. The two photons have different energies: This is obviously correct, since 1 keV is much greater than 3 eV. The frequency of the X ray photon is higher than the frequency of the visible light photon: This follows directly from E = h * f and the fact that the X ray photon has higher energy.

Common Pitfalls:
Students sometimes mistakenly think that higher energy electromagnetic waves must travel faster in vacuum. This confusion arises from everyday experience with mechanical objects where more energy can mean higher speed. However, for photons in vacuum, speed is fixed at c regardless of energy. Energy differences change frequency and wavelength, not speed. Remember that all electromagnetic waves share the same speed in vacuum, and differences appear only in media where refractive index varies with wavelength.

Final Answer:
The incorrect statement is that the speeds of the two photons in vacuum are different.