Light of two different frequencies whose photons have energies 1eV

⇦

⇨

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speed of emitted electrons will be

Options

(a) 1 : 4
(b) 1 : 2
(c) 1 : 1
(d) 1 : 5

Correct Answer:

1 : 2

Explanation:

The maximum kinetic energy of emitted electrons is given by
K.E = WorkFunction(₀) – WorkFunction(₁)
K.E₁ = 1 eV – 0.5 eV = 0.5 eV
K.E₂ = 2.5 eV – 0.5 eV = 2.0 eV
K.E₁ / K.E₂ = 0.5 eV / 2 eV = 1/4

KE = mv² /2
v₁ / v₂ = √1/4 = 1/2

Related Questions:

1. Wavefront is the locus of all points, where the particles of the medium
2. A magnifying glass of focal length 5 cm is used to view an object by a person
3. A string in musical instrument is 50 cm long and its fundamental frequency is 800 Hz.
4. A car wheel is rotated to uniform angular acceleration about its axis
5. A hollow sphere of charge does not produce an electric field at any

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