Question

calculate the energy of electron in its ground state in a one dimension box having the length of 0.2 nm (h=6.6×10^-34 J sec ,m=9.1×10^-31 kg)
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Answer 1

Answer:

sorry i don't know and thank me later on i am a kid

Answer 2

Answer:

the energy of an electron in its ground state in a one-dimension box having a length of 0.2 nm is 1.5 × 10⁻¹⁸ J.

Explanation:

Given,

The length of the box is (L) = 0.2 nm = 2 × 10⁻¹⁰m.

To find,

The energy of an electron in its ground state in a one-dimensional box.

Concept,

The energy of an 'n' energy levels of a one-dimensional box with length L is given by:

$E_n = \frac{n^2h^2}{8mL^2}$....(1)

Where,

n = number of energy levels = 1, 2, 3...

h = Planck's constant = 6.626 × 10⁻³⁴m² Kg s⁻¹

m = mass of the particle

L = length of the box in the direction of the particle oscillating.

Calculation,

Substituting h = 6.626 × 10⁻³⁴m² Kg s⁻¹, m = 9.1 × 10⁻³¹ Kg, n = 1, and L = 2 × 10⁻¹⁰m. in equation (1)

We get:

$E_1 = \frac{1^2 \times (6.626 \times 10^{-34})^2Kg^2m^4s^{-2}}{8\times 9.1 \times 10^{-31}Kg \times (2 \times 10^{-10})^2 m}$

E₁ = 1.5 × 10⁻¹⁸ J

Therefore, the energy of an electron in its ground state in a one-dimension box having a length of 0.2 nm is 1.5 × 10⁻¹⁸ J.

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