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Question
The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?
(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
Let `lambda` be the initial wavelength, V be the initial potential, `lambda^'`be the new wavelength and V' be the new operating voltage when the operating voltage is increased in the X-ray tube.
Given :-
`lambda^' = lambda - 26 "pm"`
V = 1.5 V
Energy (E) is given by
`E = (hc)/lambda`
`⇒ eV = (hc)/lambda`
Here,
h = Planck's constant
c = Speed of light
`lambda` = Wavelength of light
V = Operating potential
`therefore lambda = (hc)/(eV)`
`⇒ lambdaV = lambda^'V^' ............[∵ lambda ∝ 1/V]`
`⇒ lambda V = (lambda - 26) xx 1.5 V`
`⇒ 0.5lambda = 26 xx 1.5`
`⇒ lambda = 26 xx 3`
`⇒ lambda = 78 "pm"`
Hence, the initial wavelength is `78 xx 10^-12 "m".`
Now, the operating voltage (V) is given by
`V = (hc)/(elambda)`
`⇒ V = (6.63 xx 10^-34 xx 3 xx 10^8)/(1.6 xx 10^-19 xx 78 xx 10^-12)`
`⇒ V = 0.15937 xx 10^5`
`⇒ V = 15.9 "kV"`
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