An electron is in the n = 3 orbit of a hydrogen-like atom of atomic number Z. A neutron with thermal energy kᴮT has the same de Broglie wavelength as the electron. If T = Z²h²/(απ²a₀²mᴺkᴮ), find α.
Physics · Circular Motion
JEE Advanced 2025 Paper 1

Question

An electron is in the n = 3 orbit of a hydrogen-like atom of atomic number Z. A neutron with thermal energy kᴮT has the same de Broglie wavelength as the electron. If T = Z²h²/(απ²a₀²mᴺkᴮ), find α.

Physics · Circular Motion

JEE Advanced 2025 Paper 1

G1 Physics · Accepted Answer

For the n-th Bohr orbit, pₑrₙ = nh/(2π), rₙ = n²a₀/Z. Hence pₑ = Zh/(2πna₀). For n = 3: pₑ = Zh/(6πa₀). Equal de Broglie wavelength means pₙ = pₑ. Given neutron thermal energy kᴮT = pₙ²/(2mᴺ), T = pₑ²/(2mᴺkᴮ) = Z²h²/(72π²a₀²mᴺkᴮ). Comparing with the given form gives α = 72.