If the uncertainly in the position of an electron is $10^{- 10} m$ , then the value of uncertainty in its momentum will be __________ $k g m s^{- 1}$ . $(h = 6.62 \times 10^{- 34} J - s)$ .

1.05 \times 10^{- 24}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.03 \times 10^{- 24}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.06 \times 10^{- 24}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.08 \times 10^{- 24}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1.05 \times 10^{- 24}$
Given : $h = 6.62 \times 10^{- 34} J . s$ $Δ x = 10^{- 10} m$
value of uncertainty in its momentum $Δ p = \frac{h}{2 π Δ x}$
$⟹ Δ p = \frac{6.62 \times 10^{- 34}}{2 \times 3.142 \times 10^{- 10}} = 1.05 \times 10^{- 24} k g . m / s$

Suggest Corrections

Similar questions

The uncertainty in momentum of an electron is $1 \times 10^{- 5} k g - m / s$ . The uncertainty in its position will be ( $h = 6.62 \times 10^{- 34} k g - m^{2} / s$ )

Q. The momentum of a particle with de-Broglie wavelength

1 \circ A

is:

(h = 6.62 \times 10^{- 34} J s)

Q. The uncertainty in the momentum of an electron is

10^{- 6} k g m s^{- 1}

. What will be the uncertainty in its position? [

h = 6.6 \times - 10^{- 34} J s

]

Q. The momentum of a particle with de Broglie wavelength

2 \circ A

is:

(Planck's constant, h = 6.62 \times 10^{- 34} J s)

Q. Calculate the uncertainity in velocity of a cricket ball of mass 150 g if the uncertainty in its position is of the order of

1 o A .

(h = 6.6 \times 10^{- 34} k g m^{2} s^{- 1})

Join BYJU'S Learning Program

If the uncertainly in the position of an electron is 10−10m, then the value of uncertainty in its momentum will be __________ kgms−1. (h=6.62×10−34J−s).

The correct option is A 1.05×10−24Given : h=6.62×10−34 J.s Δx=10−10 mvalue of uncertainty in its momentum Δp=h2πΔx⟹ Δp=6.62×10−342×3.142×10−10=1.05×10−24 kg.m/s

If the uncertainly in the position of an electron is $10^{- 10} m$ , then the value of uncertainty in its momentum will be __________ $k g m s^{- 1}$ . $(h = 6.62 \times 10^{- 34} J - s)$ .

The correct option is A $1.05 \times 10^{- 24}$
Given : $h = 6.62 \times 10^{- 34} J . s$ $Δ x = 10^{- 10} m$
value of uncertainty in its momentum $Δ p = \frac{h}{2 π Δ x}$
$⟹ Δ p = \frac{6.62 \times 10^{- 34}}{2 \times 3.142 \times 10^{- 10}} = 1.05 \times 10^{- 24} k g . m / s$