Question

The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30 pm, find the electric field between the cathode and the target.

Answer 1

Explanation:

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Answer 2

The electric field between the cathode and the target is 27.6 kV/m

Explanation:

In the question, it is given that the distance between the target and the filament in the X-ray tube, d = 1.5 m.

• Cut-off wavelength, λ = 30 pm
• Energy, E is shown as E = hcλ

• where,
• h = Constant of Planck
• c = light’s speed
• λ = light’s wavelength  

• So, we have E = 1242 eV - nm30 × 10-3

• ⇒E = 1242 × 10-930 × 10-12 ⇒E = 41.4 × 103 eV  

• Now, Electric field = Vd = 41.4 × 1031.

• =27.6 × 103 V/m      

• =27.6 kV/m