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Heisenberg uncertainty principle calculation (position, velocity or mass)

This type of question, asked 4 times: 2021 Paper 26, 27 Jun 2022, 29 Jul 2022, 9 Apr 2024. All 4 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 26

    55
    A ball weighing 10 g10\text{ }g is moving with a velocity of 90 ms−190{\text{ }ms}^{- 1}. If the uncertainty in its velocity is 5%5\%, then the uncertainty in its position is x×10−33 mx \times10^{- 33}\text{ }m. (Rounded off to the nearest integer) [Given h=6.63×10−34Jsh = 6.63 \times10^{- 34}Js]

  2. Question 2

    27 Jun 2022

    52
    If the uncertainty in velocity and position of a minute particle in space are, 2.4×10−26( ms−1)2.4 \times10^{- 26}\left( {\text{ }ms}^{- 1} \right) and 10−7( m)10^{- 7}(\text{ }m) respectively. The mass of the particle in gg is (Nearest integer) (Given: h=6.626×10−34Jsh = 6.626 \times10^{- 34}Js )

  3. Question 3

    29 Jul 2022

    53
    The minimum uncertainty in the speed of an electron in an one dimensional region of length 2a02a_{0} (Where a0=a_{0}= Bohr radius  52.9pm)\left. \ 52.9pm \right) is ___ kms−1kms^{- 1}. (Given: Mass of electron =9.1×10−31 kg= 9.1 \times10^{- 31}\text{ }kg, Planck's constant  h=6.63×10−34Js)\left. \ h= 6.63 \times10^{- 34}Js \right)

  4. Question 4

    9 Apr 2024

    147
    Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter 10−15 m10^{- 15}\text{ }m is _____ ×109 ms−1\times10^{9}{\text{ }ms}^{- 1} (nearest integer) [Given: mass of electron =9.1×10−31 kg= 9.1 \times10^{- 31}\text{ }kg, Plank's constant (h)=6.626×10−34Js(h) = 6.626 \times10^{- 34}Js ] (Value of π=3.14\pi = 3.14 )