An electron is confined to move in a one dimensional potential well of length 5Å. Find the quantized energy values for the lowest energy state.
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 = ?
Answer:
The quantized energy values for the lowest energy state is 2.6hp
Explanation:
Given
Force = m g = 500 kg x 9.8 m/ s²=4900 N
Force = m g = 500 kg x 9.8 m/ s²=4900 N Velocity = 0.4 m/s
Force = m g = 500 kg x 9.8 m/ s²=4900 N Velocity = 0.4 m/s
Now, power=Force x velocity
Force = m g = 500 kg x 9.8 m/ s²=4900 N Velocity = 0.4 m/s
Now, power=Force x velocity=4900 x 0.4-1960 W
Force = m g = 500 kg x 9.8 m/ s²=4900 N Velocity = 0.4 m/s
Now, power=Force x velocity=4900 x 0.4-1960 W
Now 1 W = 0.00134 hp
Force = m g = 500 kg x 9.8 m/ s²=4900 N Velocity = 0.4 m/s
Now, power=Force x velocity=4900 x 0.4-1960 W
Now 1 W = 0.00134 hp
So, 1960 W-1960 x 0.00134=2.6 hp
To learn more about quantized energy refer:
brainly.in/question/51620389
brainly.in/question/20408014
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