Light of wavelength 500 nm is incident on a metal with work function 2.28 eV.

⇦

⇨

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.

Related Questions:

1. If two charges +4e and +e are at a distance x apart, then at what distance charge q
2. A parallel plate capacitor has 91 plates, all are identical and arranged with same
3. Volume of gas becomes four times if temperature becomes
4. In p- type semiconductor, the acceptor level lies
5. A body moves with a uniform acceleration a and zero initial velocity

Topics: Dual Nature of Matter and Radiation (150)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends