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Hrishant Goswami Grade: 10
        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. 
3 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
										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.
3 years ago
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