The uncertainty in the velocity of particle of mass 6.626 × 10-28 Kg is 10-6 m / sec . What is the uncertainty in its position in nm ?

Question

The uncertainty in the velocity of particle of mass 6.626 × 10-28 Kg is 10-6 m / sec . What is the uncertainty in its position in nm ? (A) 1 / 2 π (B) 2.5 / π (C) 4 / π (D) 1 / 4 π

Tardigrade Admin · Accepted Answer

According to Heisenberg's uncertainty principle

m Δ v \times Δ x \geq \frac{h}{4 π}

m = 6.626 \times 1 0^{- 28} k g

Δ v = 1 0^{- 6} m / sec

h = 6.626 \times 1 0^{- 34} J s

6.626 \times 1 0^{- 28} \times 1 0^{- 6} \times Δ x = \frac{6.626 \times 1 0 ^{- 34}}{4 π}

Δ x = \frac{6.626 \times 1 0 ^{- 34}}{4 π \times 6.626 \times 1 0 ^{- 34}} = \frac{1}{4 π}

Q. The uncertainty in the velocity of particle of mass $6.626 \times 1 0^{- 28} K g$ is $1 0^{- 6} m / sec .$ What is the uncertainty in its position in $nm ?$

$1/2 π$

$2.5/ π$

$4/ π$

$1/4 π$

Q. The uncertainty in the velocity of particle of mass 6.626×10−28Kg is 10−6m/sec. What is the uncertainty in its position in nm?

1/2π

2.5/π

4/π

1/4π

According to Heisenberg's uncertainty principle mΔv×Δx≥4πh​ m=6.626×10−28kg Δv=10−6m/sec h=6.626×10−34Js 6.626×10−28×10−6×Δx=4π6.626×10−34​ Δx=4π×6.626×10−346.626×10−34​=4π1​

Q. The uncertainty in the velocity of particle of mass $6.626 \times 1 0^{- 28} K g$ is $1 0^{- 6} m / sec .$ What is the uncertainty in its position in $nm ?$

$1/2 π$

$2.5/ π$

$4/ π$

$1/4 π$