A microscope using suitable photons is employed to locate an electron in an atom within the distance of $0.1^{o} A$ .

What is the Uncertainty involved in the measurement of its velocity?

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Answer 1

$Δ x . Δ p \geq \frac{h}{4 π}$ , $Δ p = m Δ v$ ,

$Δ v = \frac{Δ p}{m}$

$∴ Δ v \geq \frac{h}{4 π m Δ x}$

$\Rightarrow Δ v \geq \frac{6 . 6 2 6 \times 1 0 ^{- 34}}{4 \times 3 . 1 4 \times 9 . 1 1 \times 1 0 ^{- 31} \times 0 . 1 \times 1 0 ^{- 10}}$

$\Rightarrow Δ v \geq 5.79 \times 1 0^{6} m / s$

Answer 2

Δ x . Δ p \geq \frac{h}{4 π}

,

Δ p = m Δ v

,

Δ v = \frac{Δ p}{m}

∴ Δ v \geq \frac{h}{4 π m Δ x}

\Rightarrow Δ v \geq \frac{6 . 6 2 6 \times 1 0 ^{- 34}}{4 \times 3 . 1 4 \times 9 . 1 1 \times 1 0 ^{- 31} \times 0 . 1 \times 1 0 ^{- 10}}

\Rightarrow Δ v \geq 5.79 \times 1 0^{6} m / s

A microscope using suitable photons is employed to locate an electron in an atom within the distance of $0.1^{o} A$ .

What is the Uncertainty involved in the measurement of its velocity?

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A microscope using suitable photons is employed to locate an electron in an atom within a distance of $0.1 \overset{˚}{A}$ . What is the uncertainty involved in the measurement of its velocity? ( h = 6.626 $\times 1 0^{- 34}$ Js, mass of electron = 9.1 $\times 1 0^{- 31}$ kg)

The mass of electron is $9.11 \times 1 0^{- 31}$ kg.Planck's constant is $6.626 \times 1 0^{- 34}$ Js then the uncertainty involved in the measurement of velocity within a distance of 0.1 $\overset{˚}{A}$ is :

If uncertainty in the measurement of position and momentum are equal, then uncertainty in the measurement of velocity is equal to:

How can you determine the size of hydrogen atom using uncertainty principle potential energy of an electron $V = \frac{- e}{4 Π E _{0} a}$ , were 'a' is the distance of electron from the nucleus.

Can you locate the electron within $0.005 n m$ ?

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A microscope using suitable photons is employed to locate an electron in an atom within the distance of 0.1oA. What is the Uncertainty involved in the measurement of its velocity?

Δx.Δp≥4πh​, Δp=mΔv, Δv=mΔp​ ∴Δv≥4πmΔxh​ ⇒Δv≥4×3.14×9.11×10−31×0.1×10−106.626×10−34​ ⇒Δv≥5.79×106m/s

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Similar questions

A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1A˚. What is the uncertainty involved in the measurement of its velocity? ( h = 6.626 ×10−34 Js, mass of electron = 9.1 ×10−31 kg)

The mass of electron is 9.11×10−31 kg.Planck's constant is 6.626×10−34 Js then the uncertainty involved in the measurement of velocity within a distance of 0.1 A˚ is :

If uncertainty in the measurement of position and momentum are equal, then uncertainty in the measurement of velocity is equal to:

How can you determine the size of hydrogen atom using uncertainty principle potential energy of an electron V=4ΠE0​a−e​, were 'a' is the distance of electron from the nucleus.

Can you locate the electron within 0.005nm?

More From Chapter

Structure of Atom

Shortcuts & Tips

Diagram set

Memorization tricks

Problem solving tips

Cheatsheets

Common Misconceptions

Mindmap

Practice more questions

2 Qs

2 Qs

2 Qs

4 Qs

109 Qs

701 Qs

A microscope using suitable photons is employed to locate an electron in an atom within the distance of $0.1^{o} A$ .

What is the Uncertainty involved in the measurement of its velocity?

A microscope using suitable photons is employed to locate an electron in an atom within a distance of $0.1 \overset{˚}{A}$ . What is the uncertainty involved in the measurement of its velocity? ( h = 6.626 $\times 1 0^{- 34}$ Js, mass of electron = 9.1 $\times 1 0^{- 31}$ kg)

The mass of electron is $9.11 \times 1 0^{- 31}$ kg.Planck's constant is $6.626 \times 1 0^{- 34}$ Js then the uncertainty involved in the measurement of velocity within a distance of 0.1 $\overset{˚}{A}$ is :

How can you determine the size of hydrogen atom using uncertainty principle potential energy of an electron $V = \frac{- e}{4 Π E _{0} a}$ , were 'a' is the distance of electron from the nucleus.

Can you locate the electron within $0.005 n m$ ?