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
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
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