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Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is
Options
(a) ≥2.8×10⁻⁹ m
(b) ≤2.8×10⁻¹² m
(c) < 2.8×10⁻¹⁰ m
(d) <2.8×10⁻⁹ m
Correct Answer:
≥2.8×10⁻⁹ m
Explanation:
Given : Work function ɸ of metal = 2.28 eV
Wavelength of light λ = 500 nm = 500 x 10⁻⁹ m
KE(max) = (hc / λ) – ɸ
KE(max) = [(6.6 × 10⁻³⁴ × 3 × 10⁸) / 5 × 10⁻⁷] – 2.82 = 2.48 – 2.28 = 0.2 eV
λ(min) = h / p = h / √[2m KE(max)] = (20/3) × 10⁻³⁴ / √(2 × 9 × 10⁻³¹ × 0.2 × 1.6 × 10⁻¹⁹)
λ(min) = (25 / 9) × 10⁻⁹ = 2.80 × 10⁻⁹ nm
Therefore, λ ≥ 2.8 × 10⁻⁹ nm.
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- One coolie takes 1 minute to raise a suitcase through a height of 2 m
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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