Physics, asked by maha221, 1 month ago

An electron is bound by a potential which closely approaches an infinite square well of width 0.2nm .calculate the lowest three permissible quantum energies the electron can have​

Answers

Answered by kkhatiaxomiyaswag
1

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

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

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2. Calculate the velocity and kinetic energy of an electron of wavelength 1.66 × 10 –10 m.

Sol: Wavelength of an electron (λ) = 1.66 × 10–10 m

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To calculate KE:

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3. An electron is bound in one-dimensional infinite well of width 1 × 10–10 m. Find the energy values in the ground state and first two excited states.

Sol: Potential well of width (L) = 1 × 10–10 m

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For ground state n = 1,

E2 = 4E1 = 2.415 × 10−17 J

= 150.95 eV

E3 = 9E1 = 5.434 × 10−17 J

= 339.639 eV.

4. An electron is bound in one-dimensional box of size 4 × 10–10 m. What will be its minimum energy?

Sol: Potential box of size (L) = 4 × 10–10 m

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5. An electron is moving under a potential field of 15 kV. Calculate the wavelength of the electron waves.

Sol: V = 15 × 103 V λ = ?

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6. Find the least energy of an electron moving in one-dimensional potential box (infinite height) of width 0.05nm.

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7. A quantum particle confined to one-dimensional box of width ‘a’ is known to be in its first excited state. Determine the probability of the particle in the central half.

Sol: Width of the box, L = a

First excited state means, n = 2

Probability at the centre of the well, P2 (L/2) = ?

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The probability of the particle at the centre of the box is zero.

8. 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.

Sol: One-dimensional potential well of width, L = 3 × 10–10 m

Electron is present in ground state, so n = 1

E1 = ?

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9. Calculate the de Brogile wavelength of neutron whose kinetic energy is two times the rest mass of electron (given mn = 1.676 × 10–27 kg, me = 9.1 × 10–31 kg, C = 3 × 10 8 m/s and h = 6.63 × 10–34 J.S).

Sol: Kinetic energy of neutron, images

where mn = mass of neutron

me = mass of an electron

de Brogile wavelength of neutron, λn = ?

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10. An electron is confined to a one-dimensional potential box of length 2 Å. Calculate the energies corresponding to the second and fourth quantum states (in eV).

Sol: Length of the one-dimensional potential box, L = 2Å = 2 × 10–10 m

Energy of electron in nth level,

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Energy corresponding to second and fourth quantum states is:

E2 = 22E1 = 4 × 9.43 eV = 37.72 eV

and

E4 = 42E1 = 16 × 9.43 eV = 150.88 eV

Answered by shreyasharma4809
2

Explanation:

Length of the one-dimensional potential box, L = 2Å = 2 × 10–10 m

Energy of electron in nth level, 

Energy corresponding to second and fourth quantum states is:

E2 = 22E1 = 4 × 9.43 eV = 37.72 eV

and

E4 = 42E1 = 16 × 9.43 eV = 150.88 eV

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