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. The Boolean expression P+ P ̅ Q, where P and Q are the inputs of the logic
2. The purpose of using heavy water in nuclear reactor is
3. Small drops of the same size are charges to V volt each. If n such drops coalesce
4. If Q, E and W denote respectively the heat added, change in internal energy
5. A plane electromagnetic wave of frequency 20 MHz travels through a space along

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