Question

lowest energy of an electron trapped in a potential well 38 ev .Calculate the width of the well

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Answer 1

Calculate the wavelength associated with an electron with energy 2000 eV.

Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J



Answer 2

The width of the potential well is  2.236 x 10^29 Angstroms.

Given :

Lowest energy = 38eV

To find :

The width of the well.

Solution :

The energy of an electron trapped in a potential well is given by:

$E = (h^2) / (8 * m * L^2)$

Rearranging the equation,

$L =\sqrt{ (h^2) / (8 * m * E)}$

Substituting the given values,

L = sqrt((6.63 x 10^-34 J-s)^2 / (8 * 9.1 x 10^-31 kg * 38 eV * 1.6 x 10^-19 C))

$L = \sqrt{(6.63 * 10^-34)^2 / (8 * 9.1 x 10^-31 * 38 * 1.6 x 10^-^1^9)}$

$L = \sqrt{(4.12 * 10^-6^8 / (8.208 x 10^-^3^0)}$

$L = \sqrt{(5.00 * 10^3^8)}$

$L = 2.236 * 10^1^9$

$L = 2.236 * 10^1^9 * 10^1^0 A/m$

$L = 2.236 * 10^2^9$

So, the width of the potential well is  2.236 x 10^29 Angstroms.

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