Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter m is ______ (nearest integer)

Question

Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter $10^{- 15}$ m is ______ $\times 10^{9} {ms}^{- 1}$ (nearest integer) [Given: mass of electron = $9.1 \times 10^{- 31}$ kg, Plank's constant $(h) = 6.626 \times 10^{- 34}$ Js] (Value of $π = 3.14$ ) [2024]

Smart Achievers · Accepted Answer

(58) $Δ x \times m Δ v \geq \frac{h}{4 π}$

$Δ x = 10^{- 15} m$

Substituting other values:

$10^{- 15} \times 9.1 \times 10^{- 31} \times Δ v \geq \frac{6.626 \times 10^{- 34}}{4 \times 3.14}$

$Δ v \geq 58 \times 10^{9} m / s$

Solution

Ans.
(58)    $Δ x \times m Δ v \geq \frac{h}{4 π}$

   $Δ x = 10^{- 15} m$

  Substituting other values:

$10^{- 15} \times 9.1 \times 10^{- 31} \times Δ v \geq \frac{6.626 \times 10^{- 34}}{4 \times 3.14}$

   $Δ v \geq 58 \times 10^{9} m / s$

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Solution

Q. 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 m is ______ ×109ms-1 (nearest integer) [Given: mass of electron = 9.1×10-31 kg, Plank's constant (h)=6.626×10-34 Js] (Value of π=3.14) [2024]

Ans. (58) Δx×mΔv≥h4π Δx=10-15m Substituting other values: 10-15×9.1×10-31×Δv≥6.626×10-344×3.14 Δv≥58×109m/s

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Ans.
(58) $Δ x \times m Δ v \geq \frac{h}{4 π}$

$Δ x = 10^{- 15} m$

Substituting other values:

$10^{- 15} \times 9.1 \times 10^{- 31} \times Δ v \geq \frac{6.626 \times 10^{- 34}}{4 \times 3.14}$

$Δ v \geq 58 \times 10^{9} m / s$