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The line width of a spectral line of 600 mil wavelength emitted from a level of a lifetime of 10-8 sec, is__________×10-15 m (up to two decimal places)
Correct answer is between '17,20'. Can you explain this answer?
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The line width of a spectral line of 600 mil wavelength emitted from a...
Since, 
For λ = 600 nm, the line width is given by
For λ = 600 nm, the line width is given by
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The line width of a spectral line of 600 mil wavelength emitted from a...
The line width of a spectral line can be determined using the uncertainty principle, which states that there is a fundamental limit to the precision with which certain pairs of physical properties of a particle can be known simultaneously. In the case of a spectral line, the uncertainty in energy and time are related.
The uncertainty principle can be expressed as:
ΔE * Δt ≥ h/4π
Where ΔE is the uncertainty in energy, Δt is the uncertainty in time, and h is the Planck's constant.
In this case, we are given the lifetime of the level as 10^(-8) sec. Since the uncertainty in time represents the lifetime of the level, we can take Δt = 10^(-8) sec.
The energy of a photon is given by the equation:
E = hc/λ
Where h is Planck's constant, c is the speed of light, and λ is the wavelength of the spectral line.
Rearranging the equation, we can express the uncertainty in energy as:
ΔE = hΔc/λ^2
Where Δc is the uncertainty in the speed of light.
Substituting the values into the uncertainty principle equation:
hΔc/λ^2 * 10^(-8) ≥ h/4π
Simplifying:
Δc/λ^2 ≥ 1/(4π * 10^8)
The line width is given by the uncertainty in the speed of light, Δc. Rearranging the equation:
Δc ≥ λ^2/(4π * 10^8)
Substituting the wavelength of 600 mil (600 * 10^(-9)):
Δc ≥ (600 * 10^(-9))^2/(4π * 10^8)
Calculating the line width:
Δc ≥ 3.54 * 10^(-24) m^2/s
Therefore, the line width of the spectral line with a wavelength of 600 mil emitted from a level with a lifetime of 10^(-8) sec is approximately 3.54 * 10^(-24) m^2/s.
The uncertainty principle can be expressed as:
ΔE * Δt ≥ h/4π
Where ΔE is the uncertainty in energy, Δt is the uncertainty in time, and h is the Planck's constant.
In this case, we are given the lifetime of the level as 10^(-8) sec. Since the uncertainty in time represents the lifetime of the level, we can take Δt = 10^(-8) sec.
The energy of a photon is given by the equation:
E = hc/λ
Where h is Planck's constant, c is the speed of light, and λ is the wavelength of the spectral line.
Rearranging the equation, we can express the uncertainty in energy as:
ΔE = hΔc/λ^2
Where Δc is the uncertainty in the speed of light.
Substituting the values into the uncertainty principle equation:
hΔc/λ^2 * 10^(-8) ≥ h/4π
Simplifying:
Δc/λ^2 ≥ 1/(4π * 10^8)
The line width is given by the uncertainty in the speed of light, Δc. Rearranging the equation:
Δc ≥ λ^2/(4π * 10^8)
Substituting the wavelength of 600 mil (600 * 10^(-9)):
Δc ≥ (600 * 10^(-9))^2/(4π * 10^8)
Calculating the line width:
Δc ≥ 3.54 * 10^(-24) m^2/s
Therefore, the line width of the spectral line with a wavelength of 600 mil emitted from a level with a lifetime of 10^(-8) sec is approximately 3.54 * 10^(-24) m^2/s.
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The line width of a spectral line of 600 mil wavelength emitted from a level of a lifetime of 10-8sec,is__________×10-15 m (up to two decimal places)Correct answer is between '17,20'. Can you explain this answer?
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