Question

Q. The measurement of the electron position is associated with an uncertainty in momentum which is equal to 1X10−18gcms−1 . The uncertainty in electron velocity is

Answer 1

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Uncertainity in momentum, ∆p=1×10-18gcm/sNow we know that, p = mvor ∆p=m∆vTherefore, ∆v=∆pmm = 9×10-28gso, ∆v=1×10-18 9×10-28= =0.1111×1010cm/sor, ∆v=11.11×108cm/s

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Answer 2

Hey there !

Solution:

Given Information:

• Mass of electron = 9 × 10³¹ kg => 9 × 10⁻²⁸ g
• Uncertainty in Momentum = 1 × 10⁻¹⁸ g

To find :

• Uncertainty in position of the electron

Illustration:

Formula to be used:

• Momentum ( p ) = Mass ( m ) × Velocity ( v )

=> Δ p = Δ ( m.v )

We know that mass is constant. So we get,

=> Δ p = m Δ v

So we get,

=> Δ v = Δ p / m

Substituting the values in the above equation we get,

=> Δ v = 1 × 10⁻¹⁸ / 9 × 10⁻²⁸

=> Δ v = ( 1 / 9 ) × 10⁻¹⁸⁺²⁸

=> Δ v = 0.1111 × 10¹⁰ cm/s

=> Δ v = 11.11 × 10⁸ cm/s

Therefore the uncertainty in the velocity of an electron is 11.11 × 10⁸ cm/s.

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