Physics · Dual Nature Of Radiation And Matter · NEET
Kinetic energy is K = (1/2) m v squared. Momentum is p = m v. If you multiply K by 2m you get 2 m K = m squared v squared = (m v) squared = p squared. Take the square root: p = sqrt(2 m K). Now put this into lambda = h/p and you get lambda = h / sqrt(2 m K). This works for any slow-moving particle (electron, proton, neutron, alpha).
K sits under a square root in the bottom of the fraction, so lambda is inversely proportional to sqrt(K). If you make K four times bigger, sqrt(K) becomes two times bigger, so lambda becomes half. Wavelength never grows with energy - more energy always means a shorter matter wave.
With K fixed, lambda = h / sqrt(2 m K) depends only on mass: lambda is proportional to 1/sqrt(m). The lighter particle has the longer wavelength. Example: a proton and an alpha particle at the same energy give lambda_p / lambda_alpha = sqrt(m_alpha / m_p) = sqrt(4) = 2, so the proton's wave is twice as long.
They are the same idea. When an electron is accelerated through a voltage V, its kinetic energy is K = eV. Substitute K = eV and the electron's mass and charge into lambda = h/sqrt(2mK), and it simplifies to the shortcut lambda = 12.27/sqrt(V) angstrom (V in volts). Use the shortcut only for electrons accelerated by a potential difference; use lambda = h/sqrt(2mK) when you are given energy in joules or eV directly, or for a neutron, proton, or alpha.
Square the reciprocal: (1/lambda) squared = p squared / h squared = 2 m K / h squared. So (1/lambda) squared = (2m/h squared) times K. This is a straight line y = (constant) times K passing through the origin. NEET 2024 used exactly this graph.
The graph which shows the variation of (1/lambda) squared and the kinetic energy E of a free particle (where lambda is the de Broglie wavelength of the particle) is:
A proton and an alpha-particle are accelerated from rest to the same energy. The de Broglie wavelengths lambda_p and lambda_alpha are in the ratio:
The de Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (kelvin) and mass m is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
lambda = h / sqrt(2 m K), where h is Planck's constant, m is the particle mass, and K is its kinetic energy. It is derived by putting p = sqrt(2 m K) into lambda = h/p.
No. A photon has no rest mass, so 2 m K is meaningless for it. For a photon use lambda = hc/E, where E is the photon energy. The sqrt(2mK) form is only for particles with mass like electrons, protons, neutrons and alpha particles moving much slower than light.
Keep lambda = h/sqrt(2mK) but convert K into joules first (1 eV = 1.6 times 10 to the power minus 19 J). For an electron accelerated by voltage V, since K = eV, you can instead use the shortcut lambda = 12.27/sqrt(V) angstrom with V in volts.
Yes, through kinetic energy. For a particle in thermal equilibrium, K = (3/2) k T, so lambda = h / sqrt(3 m k T). Higher temperature means larger K and a shorter matter wave.
Because K appears as sqrt(2mK) in the denominator. So lambda is proportional to 1/sqrt(K). Making K four times larger only halves lambda, not to one quarter.
