An electron is treated as a standing de Broglie wave in a one-dimensional box of length 0.50 nm with lambda = 2L/n and purely kinetic energy. Find the kinetic energy for n=3 in eV, to two decimals. Use h = 6.626x10^-34 J s, m_e = 9.1x10^-31 kg, 1 eV = 1.6x10^-19 J.

JEE Advanced levelNumerical answerDifficulty 3 of 3as148

An electron is treated as a standing de Broglie wave in a one-dimensional box of length $0.50 nm$ with $λ = 2 L / n$ and purely kinetic energy. Find the kinetic energy for $n = 3$ in eV, to two decimals. Use $h = 6.626 \times 1 0^{- 34} J s$ , $m_{e} = 9.1 \times 1 0^{- 31} kg$ , $1 eV = 1.6 \times 1 0^{- 19} J$ .

Numerical answer type, answer in eV, tolerance 0.02.

λ = 2 L /3

, so

p = h / λ = 3 h /2 L

and

K = p^{2} /2 m = 9 h^{2} / (8 m L^{2}) = 13.569 eV