If the uncertainty in velocity and position of a minute particle in space are, 2.4×10-26 ms-1 and 10-7( m) respectively. The mass of the particle in grams is (Nearest integer)(Given : h=6.626×10-34Js )

ΔV=2.4×10-26 ms-1 Δx=10-7 m ∵Δp·Δx=h4π ∵mΔVΔx=h4π ⇒m×2.4×10-26×10-7=6.626×10-344×π m=6.6269.6×π×10-1 m=0.02198 kg m=21.98gm { nearest integer }=22

If the uncertainty in velocity and position of a minute particle in space are, 2.4×10-26 ms-1 and 10-7( m) respectively. The mass of the particle in grams is (Nearest integer)(Given : h=6.626×10-34Js )

ΔV=2.4×10-26 ms-1 Δx=10-7 m ∵Δp·Δx=h4π ∵mΔVΔx=h4π ⇒m×2.4×10-26×10-7=6.626×10-344×π m=6.6269.6×π×10-1 m=0.02198 kg m=21.98gm { nearest integer }=22

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If the uncertainty in velocity and position of a minute particle in space are, $2.4 \times 10^{- 26} ({ms}^{- 1})$ and $10^{- 7} (m)$ respectively. The mass of the particle in grams is (Nearest integer)
(Given : $h = 6.626 \times 10^{- 34} Js$ )

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Detailed Solution

$ΔV = 2.4 \times 10^{- 26} {ms}^{- 1} Δx = 10^{- 7} m ∵ Δp \cdot Δx = \frac{h}{4 π} ∵ mΔVΔx = \frac{h}{4 π} \Rightarrow m \times 2.4 \times 10^{- 26} \times 10^{- 7} = \frac{6.626 \times 10^{- 34}}{4 \times π} m = \frac{6.626}{9.6 \times π} \times 10^{- 1} m = 0.02198 kg m = 21.98 gm {nearest integer} = 22$