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Question
The electron beam in a colour TV is accelerated through 32 kV and then strikes the screen. What is the wavelength of the most energetic X-ray photon?
(Use Planck constant h = 6.63 × 10-34 Js= 4.14 × 10-15 eVs, speed of light c = 3 × 108 m/s.)
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Solution
Given:-
Potential of the electron beam, V = 32 kV = 32 × 103 V
`"Energy", E = 32 xx 10^3 "eV"`
Wavelength of the X-ray photon (`lambda`) is given by
`lambda = (hc)/E`
Here,
h = Planck's constant
c = Speed of light
`therefore lambda = (hc)/E`
`⇒ lambda = (1242 "eVnm")/(32 xx 10^3)`
`⇒ lambda = 38.8 xx 10^-3 "nm"`
`⇒ lambda = 38.8 "pm"`
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