Question

What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule?

Answer 1

The maximum energy of a photon of this X-ray  is 2×10-15 J

EXPLANATION:

• In the question, it is given that  

• X-ray’s wavelength, λ = 0.10 nm

• Constant of Planck, h = 6.63 ×10-34J-s

• Light’s speed, c = 3 × 108m/s

• Minimum wavelength is shown as

• λmin = hceV

• ⇒V = hceλmin

• ⇒V = 3 × 1081.6 × 6.63 × 10-34 x 10-19 × 10-10
• ⇒V = 12.43 × 103 V = 12.4 kV

• Photon’s maximum energy, E is shown as E = hcλ

• ⇒E = 3 × 10810-10 x 6.68 × 10-34

• ⇒E = 19.89 × 10-16

• ⇒E = 1.989 × 10-15, which is approximately 2×10-15 J.