In an X-ray tube, the distance separating the cathode (filament) and the target is 1.5 m . Determine the electric field between the cathode and the target if the cutoff wavelength is given as 30 pm .

Step-by-step solution

The cutoff wavelength _ min is related to the accelerating voltage V by the equation: _ min = hc eV Substituting the values h = 6.63 10⁻³⁴ J s , c = 3 10^8 m/s , e = 1.6 10⁻¹⁹ C , and _ min = 30 10⁻¹² m : V = 6.63 10⁻³⁴ 3 10^8 1.6 10⁻¹⁹ 30 10⁻¹² V = 41.4 10^3 V = 41.4 kV The electric field E between the cathode and the target is given by: E = V d Substituting V = 41.4 kV and d = 1.5 m : E = 41.4 1.5 = 27.6 kV/m