Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (v.) and work function (W) of the metal.
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Answer:
2.92 × 10⁻¹⁹ J
Explanation:
We have given :
Threshold wavelength λ₀ = 6800 Â
λ₀ in term of m :
= > λ₀ = 6800 × 10⁻¹⁰ m
We know :
λ = c / v
= > c = v λ
= > v₀ = c / λ₀
Putting values here we get :
= > v₀ = 3 × 10⁸ / 6800 × 10⁻¹⁰ sec⁻¹
= > v₀ = 3 × 10⁸ / 68 × 10⁻⁸ sec⁻¹
= > v₀ = 4.41 × 10¹⁴ sec⁻¹
Now we also know :
Work done W₀ = h v₀ :
= > W₀ = ( 6.626 × 10⁻³⁴ × 4.41 × 10¹⁴ ) J
= > W₀ = ( 6.626 × 4.41 × 10⁻²⁰ ) J
= > W₀ = 2.92 × 10⁻¹⁹ J
Hence we get required answer.
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