An electron is confined within Delta x = 0.20 nm. Take the minimum momentum uncertainty Delta p = h/(4piDelta x) and estimate the least kinetic energy as (Delta p)^2/2m_e. Give the answer 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 3as150

An electron is confined within $Δ x = 0.20 nm$ . Take the minimum momentum uncertainty $Δ p = h / (4 π Δ x)$ and estimate the least kinetic energy as $(Δ p)^{2} /2 m_{e}$ . Give the answer 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 0.01.

Δ p = 2.636 \times 1 0^{- 25} kg m s^{- 1}

. Then

K = (Δ p)^{2} /2 m_{e} = 3.819 \times 1 0^{- 20} J = 0.239 eV