The uncertainty in momentum of an electron is 1× 10-5 kg m/s. The uncertainty in its position will be (Given, h = 6.62 × 10-34 kg m2/s)

Question

The uncertainty in momentum of an electron is 1× 10-5 kg m/s. The uncertainty in its position will be (Given, h = 6.62 × 10-34 kg m2/s) (A) 1.05 × 10-28 m (B) 1.05 × 10-26 m (C) 5.27 × 10-30 m (D) 5.25 × 10-28 m

Tardigrade Admin · Accepted Answer

According to Heisenberg's uncertainty principle

Δ p \times Δ x \geq \frac{h}{4 π}

Uncertainty in momentum

Δ p = 1 \times 1 0^{- 6} k g m / s

1 \times 1 0^{- 5} \times Δ x = \frac{6.62 \times 1 0 ^{- 34}}{4 \times \frac{22}{7}}

Δ x = \frac{6.62 \times 1 0 ^{- 34} \times 7}{1 \times 1 0 ^{- 6} \times 4 \times 22}

= 5.265 \times 1 0^{- 30} m

\approx 5.27 \times 1 0^{- 30} m

Q. The uncertainty in momentum of an electron is $1 \times 1 0^{- 5} k g m / s .$ The uncertainty in its position will be
(Given, $h = 6.62 \times 1 0^{- 34} k g m^{2} / s)$

$1.05 \times 1 0^{- 28} m$

$1.05 \times 1 0^{- 26} m$

$5.27 \times 1 0^{- 30} m$

$5.25 \times 1 0^{- 28} m$

Q. The uncertainty in momentum of an electron is 1×10−5kgm/s. The uncertainty in its position will be (Given, h=6.62×10−34kgm2/s)

1.05×10−28m

1.05×10−26m

5.27×10−30m

5.25×10−28m

According to Heisenberg's uncertainty principle Δp×Δx≥4πh​ Uncertainty in momentum Δp=1×10−6kgm/s 1×10−5×Δx=4×722​6.62×10−34​ Δx=1×10−6×4×226.62×10−34×7​ =5.265×10−30m ≈5.27×10−30m

Q. The uncertainty in momentum of an electron is $1 \times 1 0^{- 5} k g m / s .$ The uncertainty in its position will be
(Given, $h = 6.62 \times 1 0^{- 34} k g m^{2} / s)$

$1.05 \times 1 0^{- 28} m$

$1.05 \times 1 0^{- 26} m$

$5.27 \times 1 0^{- 30} m$

$5.25 \times 1 0^{- 28} m$