The position of an electron is known to within 0.50 AA. Using Delta x Delta p = h/4pi, h = 6.626x10^-34 J s and m_e = 9.1x10^-31 kg, find the minimum uncertainty in its speed in units of 10^6 m s^-1, to two decimals.

JEE Main levelNumerical answerDifficulty 2 of 3as129

The position of an electron is known to within $0.50 \overset{˚}{A}$ . Using $Δ x Δ p = h /4 π$ , $h = 6.626 \times 1 0^{- 34} J s$ and $m_{e} = 9.1 \times 1 0^{- 31} kg$ , find the minimum uncertainty in its speed in units of $1 0^{6} m s^{- 1}$ , to two decimals.

Numerical answer type, answer in 1e6 m/s, tolerance 0.01.

Δ v = h / (4 π m_{e} Δ x) = 6.626 \times 1 0^{- 34} / (4 π \times 9.1 \times 1 0^{- 31} \times 0.50 \times 1 0^{- 10}) = 1.159 \times 1 0^{6} m s^{- 1}