If uncertainities in measurement of position and momentum of an electron are equal, then the uncertainty in measurement of its velocity is $x \times 10^{12}$ . The value of x (the closest whole number value) is
( $m_{c} = 9.1 \times 10^{- 31} k g$ and $h = 6.625 \times 10^{- 34} J s$ )

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Solution

According to Heisenberg principle-
$Δ x Δ p \geq \frac{h}{4 π} \Rightarrow Δ x = Δ p = \sqrt{\frac{h}{4 π}} = 0.726 \times 10^{- 17}$
$Δ p = m Δ v \Rightarrow Δ v = \frac{Δ p}{M}$
$\Rightarrow \frac{0.726 \times 10^{- 17}}{9.1 \times 10^{- 31}} = 7.98 \times 10^{12} m s^{- 1}$

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If uncertainities in measurement of position and momentum of an electron are equal, then the uncertainty in measurement of its velocity is x×1012. The value of x (the closest whole number value) is(mc=9.1×10−31kg and h=6.625×10−34Js)

According to Heisenberg principle-ΔxΔp≥h4π⇒Δx=Δp=√h4π=0.726×10−17Δp=mΔv⇒Δv=ΔpM⇒0.726×10−179.1×10−31=7.98×1012ms−1

If uncertainities in measurement of position and momentum of an electron are equal, then the uncertainty in measurement of its velocity is $x \times 10^{12}$ . The value of x (the closest whole number value) is
( $m_{c} = 9.1 \times 10^{- 31} k g$ and $h = 6.625 \times 10^{- 34} J s$ )

According to Heisenberg principle-
$Δ x Δ p \geq \frac{h}{4 π} \Rightarrow Δ x = Δ p = \sqrt{\frac{h}{4 π}} = 0.726 \times 10^{- 17}$
$Δ p = m Δ v \Rightarrow Δ v = \frac{Δ p}{M}$
$\Rightarrow \frac{0.726 \times 10^{- 17}}{9.1 \times 10^{- 31}} = 7.98 \times 10^{12} m s^{- 1}$