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?

Answer 1

The original value of the operating voltage is 15.9 kV .

Explanation:

Let us consider that λ is the initial wavelength, λ’ is the new wavelength, V is the initial potential, and V‘ is the new operating voltage when there is the increase in the operating voltage in the X-ray tube.

It is given that:

• λ’ = λ - 26 pm
• V‘ = 1.5 V

• Energy, E is shown as E = hcλ

• ⇒eV = hcλ
• where,

• h = Constant of Planck

• c = light’s speed

• V = Operating potential

• λ = light’s wavelength  
• Therefore, λ = hceV  

• ⇒λV = λ’V’            

• Therefore λ ∝ 1V ⇒λ = 3 x 26 ⇒λ=78 pm.

• So, 78 ×10-12 m is the initial wavelength.

• Now, the operating voltage (V) is shown as

• V = hceλ ⇒ V = 0.15937 × 105

• ⇒V = 15.9 kV