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Question: Assuming an electron is confined to a 1 nm wide region. Find the uncertainty in momentum using Heise...

Assuming an electron is confined to a 1 nm wide region. Find the uncertainty in momentum using Heisenberg uncertainty principle. (Take h=6.63×10-34 J s )

A

1.05×10−25kgs−11.05 \times 10^{- 25}kgs^{- 1}

B

2.03×10−31kgs−12.03 \times 10^{- 31}kgs^{- 1}

C

3.05×10−34kgs−13.05 \times 10^{- 34}kgs^{- 1}

D

2.49×10−32kgs−12.49 \times 10^{- 32}kgs^{- 1}

Answer

1.05×10−25kgs−11.05 \times 10^{- 25}kgs^{- 1}

Explanation

Solution

: Here Δx=1\Delta x = 1nm =10−9m= 10^{- 9}m

By Heisenberg uncertainty principle

ΔxΔp=ℏ\Delta x\Delta p = \hslash

∴Δp=ℏΔx=h2πΔx\therefore\Delta p = \frac{\hslash}{\Delta x} = \frac{h}{2\pi\Delta x} (∴ℏ=h2π)\left( \therefore\hslash = \frac{h}{2\pi} \right)

=6.63×10−342×π×10−9=1.05×10−25kgms−1= \frac{6.63 \times 10^{- 34}}{2 \times \pi \times 10^{- 9}} = 1.05 \times 10^{- 25}kgms^{- 1}