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.)

Answer 1

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"`