Question

If the uncertainty in position of an electron is 4 x 10^(-10) m, the uncertainty in its momentum

Answer 1

Answer:

5×10^-5 kgm/s

Explanation:

$according \: to \: principle \: of \: uncertainty \: \\ ∆ x ∆ p \geqslant \frac{h}{4\pim}$

where,

• ∆x= uncertainty in position
• ∆p= uncertainty in momentum
• h= Planck's constant
• m=mass(here,mass of electron)

therefore,

uncertainty in momentum,∆p is given by,

$\alpha p = \frac{h}{ \alpha x4\pim} \\ = \frac{6.626 \times {10}^{ - 34} }{4 \times 3.14 \times 9.1 \times {10}^{ - 31} } \\ = 5 \times {10}^{ - 5}$

Answer 2

According to Heisenberg's uncertainty principle

∆x∆p⩾h/4 pi

Given uncertainty of position is

∆x

= 4 × {10}^{ - 10}

∆p⩾ h/(4×pi×∆x)

∆p

⩾ (6.63×{10}^{ - 34} )/(4×3.14×4 × {10}^{ - 10} )

>0.13 ×{10}^{ - 24} kg - m s^{ - 1}