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An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelength emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

Hrishant Goswami , 10 Years ago
Grade 10
anser 1 Answers
Jitender Pal

Last Activity: 10 Years ago

V = 40 KV = 40 * 10^3 V Energy = 40 * 10^3 eV Energy utilized = 70/100 *40 * 10^3 = 28 * 10^3 eV 11. V = 40 KV = 40 * 10^3 V Energy = 40 * 10^3 eV Energy utilized = 70/100 * 40 * 10^3 = 28 * 10^3 eV λ = hc/E = 1242 – ev nm/ 28* 10^3 ev ⇒ 444.35 * 10^-3 nm = 44.35 pm. For other wavelengths, E = 70% (left over energy) = 70/100 * (40 - 28)10^3 = 84 * 10^2. λ’ = hc/E = 1242/ 25.2* 10^2 = 49.2857 * 10^-2 nm = 493 pm.

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