Physics · Dual Nature Of Radiation And Matter · NEET
The cutoff wavelength comes from the continuous X-ray spectrum (bremsstrahlung), made when an electron is suddenly slowed by the target's nuclei. The shortest wavelength (highest energy photon) is made when a single electron loses ALL its kinetic energy eV in one collision, giving one photon: eV = hc/lambda-min. Since eV is fixed by the accelerating voltage alone, lambda-min = hc/(eV) does not contain the target metal at all. The target metal only decides the CHARACTERISTIC lines (the sharp peaks), not the cutoff.
Cutoff wavelength (lambda-min) is the sharp left edge of the continuous spectrum. It depends only on the tube voltage V. Characteristic wavelengths are sharp peaks sitting on top of the continuous curve; they depend only on the target metal (its electron energy levels), not on V. For NEET: if the question mentions voltage, use lambda-min = hc/(eV). If it mentions the target metal or K-alpha lines, that is characteristic X-rays.
It decreases. Since lambda-min = hc/(eV), lambda-min is inversely proportional to V (lambda-min is proportional to 1/V). Higher voltage gives electrons more kinetic energy, so the top photon is more energetic, meaning shorter wavelength. This exact relation (lambda-min proportional to 1/V) was asked in NEET 2023.
First find the electron's kinetic energy from its de Broglie wavelength. Momentum p = h/lambda, so kinetic energy E = p squared / (2m) = h squared / (2 m lambda squared). At cutoff all of E becomes one photon: E = hc/lambda-0. Setting them equal: h squared/(2 m lambda squared) = hc/lambda-0, which gives lambda-0 = 2 m c lambda squared / h. This is the NEET 2016 answer.
No, they are opposite processes. Photoelectric effect: a photon comes IN and knocks an electron OUT (hf = work function + KE). Duane-Hunt / X-ray production: an electron comes IN and a photon is made OUT (eV = hc/lambda-min at cutoff). Both use energy conservation and the same constant hc, so the maths looks similar, but the direction of energy transfer is reversed.
Electrons of mass m with de Broglie wavelength lambda fall on the target in an X-ray tube. The cutoff wavelength (lambda-0) of the emitted X-ray is:
The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
lambda-min = hc/(eV), where h is Planck's constant, c the speed of light, e the electron charge and V the tube voltage. In easy units: lambda-min (in angstrom) = 12400 / V(volts), or lambda-min (in nm) = 1240 / V(volts).
No. The cutoff (minimum) wavelength depends only on the accelerating voltage V. The target material only affects the characteristic sharp lines, not the short-wavelength cutoff.
It is bremsstrahlung: X-rays made when incoming electrons are slowed down and deflected by the target nuclei. Different electrons lose different amounts of energy, giving a continuous range of wavelengths above lambda-min.
An electron can give at most all its kinetic energy eV to one photon (giving the shortest wavelength). It can give less, producing longer wavelengths, so there is no upper limit; the spectrum fades out gradually on the long-wavelength side.
Use hc = 1240 eV.nm (or 12400 eV.angstrom). This lets you write lambda-min in nm as 1240 divided by the voltage in volts directly.
