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

Introduction / Context:
Different regions of the electromagnetic spectrum, such as X rays and visible light, differ in energy, frequency, and wavelength but share some common properties in vacuum. This question compares an X-ray photon and a visible photon, asking which statement about their properties is not correct. Understanding the relationships between energy, frequency, wavelength, and speed for photons is fundamental in modern physics and electromagnetic theory.

Given Data / Assumptions:
• X-ray photon energy: 1 keV (1000 eV). • Visible photon energy: 3 eV. • Both are electromagnetic waves and photons. • We consider their behaviour in vacuum.

Concept / Approach:
For a photon, energy E is related to frequency f by E = h * f, where h is Planck constant. Because 1 keV is much greater than 3 eV, the X-ray photon has much higher frequency. Wave speed v in a given medium is related to frequency and wavelength by v = f * lambda. In vacuum, all electromagnetic waves travel at the same speed, the speed of light c. Therefore, if energy and frequency increase, wavelength must decrease to keep v constant. This means X rays have shorter wavelengths than visible light, and they are generally more penetrating. The only statement that contradicts these principles is any claim that their speeds in vacuum are different.

Step-by-Step Solution:
Step 1: Compare energies: E_X-ray = 1 keV = 1000 eV, E_visible = 3 eV. So the X-ray photon has much higher energy. Step 2: Use E = h * f. Higher energy implies higher frequency. So f_X-ray > f_visible. Step 3: Use v = f * lambda. In vacuum, v = c for both photons. Step 4: With the same speed c and higher frequency, the X-ray photon must have shorter wavelength than the visible photon: lambda_X-ray < lambda_visible. Step 5: Because both are electromagnetic waves in vacuum, their speeds are identical: v_X-ray = v_visible = c. Step 6: Therefore, any statement claiming different speeds in vacuum is not correct.

Verification / Alternative check:
Electromagnetic theory and many experiments show that all electromagnetic waves travel at the same speed c in vacuum, approximately 3 * 10^8 m/s, regardless of frequency or energy. This is true for radio waves, microwaves, infrared, visible, ultraviolet, X rays, and gamma rays. Differences arise in frequency, wavelength, and interaction with matter, not in their speed in vacuum. So the idea that their speeds in vacuum are different is definitely incorrect.

Why Other Options Are Correct (and therefore not the answer):
Option a: With higher energy and frequency, X rays have shorter wavelength than visible light; this is a standard fact of the spectrum. Option b: By definition, 1 keV and 3 eV are very different energies, so the photons do have different energies. Option d: Higher energy implies higher frequency via E = h * f, so the X-ray photon frequency is higher than that of visible light. Option e: X-ray photons interact differently with matter and are generally far more penetrating than visible photons, which are easily absorbed or scattered.

Common Pitfalls:
Students sometimes incorrectly believe that higher energy photons must travel faster, but for electromagnetic waves in vacuum, speed is fixed at c. The extra energy appears as higher frequency (and shorter wavelength), not increased speed. Another confusion is mixing up behaviour in matter versus vacuum; speeds can differ in materials, but in vacuum all electromagnetic waves share the same speed.

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