Question

If the position of a 5 k eV electron is located within 2 Å, what is the percentage uncertainty in its momentum?
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Answer 1

Answer:

10^-6 this will correct answer

Answer 2

The percentage uncertainty in momentum is 2.16%

GIVEN

The position of a 5 k eV electron is located within 2 Å

TO FIND

Percentage uncertainty in the momentum.

SOLUTION

We can simply solve the above problem as follows-

Energy of electron = 5 Kev

Converting it to jules;

= 5 × 10³ × 1.6 × 10⁻¹⁹ j

= 8 × 10⁻¹⁶ j

Momentum of electron, p = √(2m Ke)

$= \sqrt{2 \times 9.1 \times {10}^{ - 31} \times 8 \times {10}^{ - 16} }$

= 1.20 × 10⁻²⁴ kg ms⁻¹

According to Heisenberg's Uncertainty principle -

$∆p∆x > \: \frac{h}{4\pi}$

Where,

∆p = uncertainty in momentum.

∆x = uncertainty in position

h = plank's constant = 6.62 × 10⁻³⁴

We know that,

1 Å = 10⁻² meter

2 Å = 2 × 10⁻² meters

So,

$Δp = \frac{6.62 \times {10}^{ - 34} \times {10}^{2} }{4 \times 3.14 \times 2}$

Δp = 2.6 × 10⁻²⁵ Kgms⁻¹

Pecentage uncertainity in momentum,

$\frac{Δp}{p} \times 100$

=

$= \frac{2.6 \times {10}^{ - 26} }{1.2 \times {10}^{ - 25} } \times 100$

= 2.16%

Hence, The percentage uncertainty in momentum is 2.16%

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