A photon of X-ray has energy of 1 keV. A photon of visible radiation has energy of 3 eV. In this context, which one of the following statements is not correct?
The wavelength of X-ray photon is less than the wavelength of visible radiation photon.
Both the photons have different energies.
The speeds of both the photons in vacuum are different.
The frequency of X-ray photon is higher than the frequency of visible radiation photon.
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CDS-1 – 2017
The energy (E), frequency (f), and wavelength (λ) of a photon are related by the equations E = hf and E = hc/λ, where h is Planck’s constant and c is the speed of light in vacuum. The speed of light in vacuum (c) is a universal constant, approximately 3 x 10⁸ m/s, and is the same for all electromagnetic radiations regardless of their energy, frequency, or wavelength.
– The energy of the X-ray photon (1 keV = 1000 eV) is significantly higher than the energy of the visible radiation photon (3 eV).
– Statement A: Since E = hc/λ, higher energy means shorter wavelength. E(X-ray) > E(Visible), so λ(X-ray) < λ(Visible). This statement is correct.
- Statement B: The energies (1000 eV and 3 eV) are clearly different. This statement is correct.
- Statement C: The speed of all electromagnetic radiation (including photons of X-rays and visible light) in vacuum is the constant 'c'. Their speeds in vacuum are the same, not different. This statement is incorrect.
- Statement D: Since E = hf, higher energy means higher frequency. E(X-ray) > E(Visible), so f(X-ray) > f(Visible). This statement is correct.