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A nucleus has a size of 10–15 m. Consider an electron bound within a nucleus. The estimated energy of this electron is of the order of
- a)1 MeV
- b)102 MeV
- c)104 MeV
- d)106 MeV
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A nucleus has a size of 1015 m. Consider an electron bound within a nu...
Explanation:
Size of Nucleus:
The size of the nucleus is given as 10^-15 m.
Electron Bound within a Nucleus:
An electron bound within a nucleus is also known as a "bound state" electron. In this state, the electron is confined to a very small space within the nucleus.
Energy of Electron:
The estimated energy of the electron can be calculated using the Uncertainty Principle. According to the Uncertainty Principle, the product of the uncertainty in position and the uncertainty in momentum is always greater than or equal to Planck's constant divided by 2π.
ΔxΔp ≥ h/2π
Where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant.
For an electron bound within a nucleus, the uncertainty in position is equal to the size of the nucleus (10^-15 m). Therefore, we can calculate the uncertainty in momentum as follows:
Δp ≥ h/2πΔx
Δp ≥ (6.626 x 10^-34 J s)/(2π x 10^-15 m)
Δp ≥ 1.054 x 10^-19 kg m/s
Using the formula for kinetic energy (K = p^2/2m), we can calculate the estimated energy of the electron:
K = (Δp)^2/2m
K = (1.054 x 10^-19 kg m/s)^2/(2 x 9.1 x 10^-31 kg)
K = 1.17 x 10^-12 J
Converting this to MeV, we get:
K = 1.17 x 10^-12 J / (1.6 x 10^-13 J/MeV)
K = 7.3 MeV
Therefore, the estimated energy of the electron bound within a nucleus is of the order of 102 MeV (approximately 7.3 MeV). Option B is the correct answer.
Size of Nucleus:
The size of the nucleus is given as 10^-15 m.
Electron Bound within a Nucleus:
An electron bound within a nucleus is also known as a "bound state" electron. In this state, the electron is confined to a very small space within the nucleus.
Energy of Electron:
The estimated energy of the electron can be calculated using the Uncertainty Principle. According to the Uncertainty Principle, the product of the uncertainty in position and the uncertainty in momentum is always greater than or equal to Planck's constant divided by 2π.
ΔxΔp ≥ h/2π
Where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant.
For an electron bound within a nucleus, the uncertainty in position is equal to the size of the nucleus (10^-15 m). Therefore, we can calculate the uncertainty in momentum as follows:
Δp ≥ h/2πΔx
Δp ≥ (6.626 x 10^-34 J s)/(2π x 10^-15 m)
Δp ≥ 1.054 x 10^-19 kg m/s
Using the formula for kinetic energy (K = p^2/2m), we can calculate the estimated energy of the electron:
K = (Δp)^2/2m
K = (1.054 x 10^-19 kg m/s)^2/(2 x 9.1 x 10^-31 kg)
K = 1.17 x 10^-12 J
Converting this to MeV, we get:
K = 1.17 x 10^-12 J / (1.6 x 10^-13 J/MeV)
K = 7.3 MeV
Therefore, the estimated energy of the electron bound within a nucleus is of the order of 102 MeV (approximately 7.3 MeV). Option B is the correct answer.
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Community Answer
A nucleus has a size of 1015 m. Consider an electron bound within a nu...
ANS= D)
EXPLANATION
E=p^2/2m
p=h/wavelength
p= 6.626×10^-34/10^-15
E=2.412×10^-7 J
but 1J=1/1.6×10^-13 mev
so E=1.50×10^6 Mev
EXPLANATION
E=p^2/2m
p=h/wavelength
p= 6.626×10^-34/10^-15
E=2.412×10^-7 J
but 1J=1/1.6×10^-13 mev
so E=1.50×10^6 Mev
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A nucleus has a size of 1015 m. Consider an electron bound within a nucleus. The estimated energy of this electron is of the order ofa)1 MeVb)102 MeVc)104 MeVd)106 MeVCorrect answer is option 'B'. Can you explain this answer?
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