Question

Find the lowest energy level and momentum of an electron in one dimensional potential well
of width 1A°​

Answer 1

Explanation:

Answer:

An electron is confined in one-dimensional potential well of width 3 × 10–10 m. Find the kinetic energy of electron when it is in the ground state. E1 = ? 9.

Explanation:

Answer 2

Given:

WIdth of the well, L = 1 A° = 1 × 10⁻¹⁰ m

Find:

The lowest energy value and momentum.

Solution:

The energy of an electron in one-dimensional potential well is given by,

$E_n = \frac{n^2 h^2}{8mL^2}$

where Eₙ = Energy

n = principal quantum number

h = planck's constant = 6.63 × 10⁻³⁴ Js

m = mass of the electron = 9.1 × 10⁻³¹ kg

L = width of the well

For lowest energy level that is, ground state, n = 1

∴  The lowest energy level is given by,

$E_1 = \frac{1^2 h^2}{8mL^2}$

$E_1 = \frac{h^2}{8mL^2}$

$E_1 = \frac{(6.63 \times 10^{-34})^2}{8(9.1\times 10^{-31})(1\times 10^{-10})^2}$

$E_1 = \frac{43.9569 \times 10^{-68}}{8(9.1\times 10^{-31})(10^{-20})}$

$E_1 = \frac{43.9569 \times 10^{-68}}{72.8 \times 10^{-31-20}}$

$E_1 = \frac{43.9569 \times 10^{-68}}{72.8 \times 10^{-51}}$

$E_1 = 0.6038 \times 10^{-68+51} \ J \\\\\ E_1 = 0.6038 \times 10^{-17} \ J$

Energy in electron-volt = $\frac{0.6038\times 10^{-17}}{1.6 \times 10^{-19}}$ = 37.74 eV

Hence, lowest energy , E₁ = 0.6038 × 10⁻¹⁷ J = 37.74 eV

Now, the relation between energy and momentum is as follows,

p = $\sqrt{2mE_1}$

Putting all values in the above equation, we get

Momentum, $p = \sqrt{2mE_1}$

                        $= \sqrt{2(9.1 \times 10^{-31})(0.6038 \times 10^-17)}$

                        $= \sqrt{10.989\times 10^{-31-17}} \\= \sqrt{10.989\times 10^{-48}}\\= \sqrt{109.89\times 10^{-49}}\\= 10.48 \times 10^{-7}\\= 1.048 \times 10^{-6} \ kg \ ms^{-1}$

Hence, the momentum of the electron is 1.048 × 10⁻⁶ kg m s⁻¹.

#SPJ2