An X-ray tube operates at 38.08 kV. The voltage is reduced to half. The difference between the Kα wavelength and the minimum continuous X-ray wavelength becomes four times its original value. Assuming the Kα line exists in both cases, find the atomic number of the target. Take Rhc = 13.6 eV.
Physics · Atomic Structure and X-rays
JEE Mains Advanced

Question

An X-ray tube operates at 38.08 kV. The voltage is reduced to half. The difference between the Kα wavelength and the minimum continuous X-ray wavelength becomes four times its original value. Assuming the Kα line exists in both cases, find the atomic number of the target. Take Rhc = 13.6 eV.

Physics · Atomic Structure and X-rays

G1 Physics · Accepted Answer

Step 1: Use λₘᵢₙ=hc/(eV) and Moseley's law for Kα. Step 2: Apply the condition that [λ(Kα)−λₘᵢₙ] becomes four times when V is halved. Solving the resulting equation gives target atomic number Z=41.