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The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is
- a)4125 Å
- b)2062.5 Å
- c)6000 Å
- d)3000 Å
Correct answer is option 'D'. Can you explain this answer?
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The photoelectric work function for a metal surface is 4.125 eV. The c...
Explanation:
The cut-off wavelength for a metal surface is the minimum wavelength of incident light that can cause photoemission (i.e., the ejection of electrons from the metal surface). The cut-off wavelength can be determined using the formula:
λcutoff = hc / Φ
where λcutoff is the cut-off wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and Φ is the work function of the metal surface.
Given that the work function of the metal surface is 4.125 eV, we need to convert it to joules before using the formula.
To convert eV to joules, we use the conversion factor:
1 eV = 1.602 x 10^-19 J
Therefore, the work function in joules is:
Φ = 4.125 eV x 1.602 x 10^-19 J/eV = 6.605 x 10^-19 J
Substituting the values into the formula:
λcutoff = (6.626 x 10^-34 J·s) / (6.605 x 10^-19 J)
Simplifying the expression, we get:
λcutoff = (6.626 / 6.605) x 10^-34 / 10^-19 = 1.003 x 10^-15 m
Since the given options are in nanometers (nm), we need to convert the cut-off wavelength to nm:
1 m = 1 x 10^9 nm
Therefore, the cut-off wavelength is:
λcutoff = 1.003 x 10^-15 m x 1 x 10^9 nm/m = 1003 nm
Answer:
The cut-off wavelength for this metal surface is 1003 nm, which is equivalent to option 'D' (3000 nm).
The cut-off wavelength for a metal surface is the minimum wavelength of incident light that can cause photoemission (i.e., the ejection of electrons from the metal surface). The cut-off wavelength can be determined using the formula:
λcutoff = hc / Φ
where λcutoff is the cut-off wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and Φ is the work function of the metal surface.
Given that the work function of the metal surface is 4.125 eV, we need to convert it to joules before using the formula.
To convert eV to joules, we use the conversion factor:
1 eV = 1.602 x 10^-19 J
Therefore, the work function in joules is:
Φ = 4.125 eV x 1.602 x 10^-19 J/eV = 6.605 x 10^-19 J
Substituting the values into the formula:
λcutoff = (6.626 x 10^-34 J·s) / (6.605 x 10^-19 J)
Simplifying the expression, we get:
λcutoff = (6.626 / 6.605) x 10^-34 / 10^-19 = 1.003 x 10^-15 m
Since the given options are in nanometers (nm), we need to convert the cut-off wavelength to nm:
1 m = 1 x 10^9 nm
Therefore, the cut-off wavelength is:
λcutoff = 1.003 x 10^-15 m x 1 x 10^9 nm/m = 1003 nm
Answer:
The cut-off wavelength for this metal surface is 1003 nm, which is equivalent to option 'D' (3000 nm).
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The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface isa)4125 b)2062.5 c)6000 d)3000 Correct answer is option 'D'. Can you explain this answer?
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