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. Charge passing through a conductor of cross-section area A=0.3 m² is given
2. A common emitter amplifier has a voltage gain of 50, an input impedance of 100Ω
3. The radioactivity of a certain material drops to 1/16 of the initial value in 2 hours.
4. A freely suspended bar magnet aligns itself in
5. 24 cells of emf 1.5 V each having internal resistance of 1 ohm are connected

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