Exactly 50 % of the X-rays emitted from a Coolidge tube are able to transmit through an aluminium foil of thickness 0.1 mm . If the potential difference between the target and the filament is raised, the fraction of the X-rays penetrating the identical foil will be

Step-by-step solution

The maximum energy of X-ray photons emitted from a Coolidge tube is given by E_ max = eV , where V is the potential difference between the target and the filament. When the potential difference V is increased, the energy of the emitted X-rays increases. These higher energy X-rays are known as hard X-rays and have a greater penetrating power. The intensity of transmitted X-rays is given by I = I₀ e^ - x , where is the absorption coefficient. As the energy of X-rays increases, the absorption coefficient of the materi