If the potential difference applied to the tube is doubled and the separation between the filament and the target is also doubled, the cutoff wavelength

Question

Philoid · Accepted Answer

The cut-off wavelength is given by:

where, λ_min is the cut-off wavelength, h is the Planck’s constant (6.62607004 × 10^-34 m² kg / s), c is the speed of light in vacuum, e is the charge of the electron and V is the potential difference applied at the ends of the tube
So, doubling the applied potential will reduce the cut-off wavelength by half.