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The threshold wavelength for photoelectric effect of a metal is 6500 Å. The work function of the metal is approximately
- a)2 eV
- b)1 eV
- c)0.1 eV
- d)3 eV
Correct answer is option 'A'. Can you explain this answer?
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The threshold wavelength for photoelectric effect of a metal is 6500 ....
Explanation:
- The threshold wavelength (λ0) for photoelectric effect of a metal is 6500 Å.
- The work function (ϕ) of the metal is given by:
hν = ϕ + KEmax
- where h is Planck's constant, ν is the frequency of incident radiation, KEmax is the maximum kinetic energy of emitted photoelectrons.
- Since λν = c, where c is the speed of light, we can write:
h(c/λ) = ϕ + KEmax
- For photoelectrons emitted at threshold frequency or wavelength, KEmax = 0. Therefore, we get:
h(c/λ0) = ϕ
- Substituting the given values, we get:
6.63 × 10-34 J.s × (3 × 108 m/s) / (6500 × 10-10 m) = ϕ
ϕ = 3.06 eV
- Therefore, the work function of the metal is approximately 3 eV.
- The closest option is A, which is 2 eV. This is because the given value of threshold wavelength is approximate and may not be exact. Also, the work function may depend on the particular sample of the metal.
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The threshold wavelength for photoelectric effect of a metal is 6500 ....
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The threshold wavelength for photoelectric effect of a metal is 6500 . The work function of the metal is approximatelya)2 eVb)1 eVc)0.1 eVd)3 eVCorrect answer is option 'A'. Can you explain this answer?
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