The uncertainty in position and velocity of the particle are 0-1 nm and 5-27- 10 -24 ms -1 respectively then find the mass of the particle h-6-625- 10 -34 Js -1-

As per Heisenberg's uncertainty principle, #8710;x #8710;p=h4π #8710;x=0.1nm=10^ 10m , #8710;p=m #8710;v and #8710;v=5.27× 10^ 24m/s ...

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Standard XII

Chemistry

Heisenberg's Uncertainty Principle

The uncertain...

Question

The uncertainty in position and velocity of the particle are $0.1 n m$ and $5.27 \times 10^{- 24} {m s}^{- 1}$ respectively then find the mass of the particle ( $h = 6.625 \times 10^{- 34} {J s}^{- 1}$ ).

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Solution

As per Heisenberg's uncertainty principle,
$Δ x Δ p = \frac{h}{4 π}$
$Δ x = 0.1 n m = 10^{- 10} m$ , $Δ p = m Δ v$ and $Δ v = 5.27 \times 10^{- 24} m / s$
$(10^{- 10} m) (5.27 \times 10^{- 24} m / s) = \frac{h}{4 π m}$
$m = \frac{6.63 \times 10^{- 34} J s}{(10^{- 10} m) (5.27 \times 10^{- 24} m / s) (4 π)}$
$m = 0.1 k g$

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The uncertainty in position and velocity of the particle are 0.1nm and 5.27×10−24ms−1 respectively then find the mass of the particle (h=6.625×10−34Js−1).

As per Heisenberg's uncertainty principle,ΔxΔp=h4πΔx=0.1nm=10−10m , Δp=mΔv and Δv=5.27×10−24m/s(10−10m)(5.27×10−24m/s)=h4πmm=6.63×10−34Js(10−10m)(5.27×10−24m/s)(4π)m=0.1kg

The uncertainty in position and velocity of the particle are $0.1 n m$ and $5.27 \times 10^{- 24} {m s}^{- 1}$ respectively then find the mass of the particle ( $h = 6.625 \times 10^{- 34} {J s}^{- 1}$ ).