How to Use NCERT Solutions Effectively
Learn the best strategies to make the most of NCERT solutions for your exam preparation.
Read MoreWelcome to the NCERT Solutions for Class 12 Physics - Chapter Dual Nature Of Radiation And Matter. This page offers a step-by-step solution to the specific question from Exercise 1, Question 2: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
Work function of caesium metal, ø0 = 2.14 eV
Frequency of light, v = 6.0 x 1014 Hz
(a) The maximum kinetic energy is given by the photoelectric effect as: K = hv - ø0
Where,
h = Planck’s constant = 6.626 × 10−34 Js
∴ K = 6.626 x 1034 x 6 x 1014 / 1.6 x 10-19 - 2.14 = 2.485 - 2.14 = 0.345 eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.
( b ) For stopping potential , V0 we can write the equation for kinetic energy as: K = eV0
∴ V0 = K/e = 0.345 x 1.6 x 10-19 / 1.6 x 10-19 = 0.345 V
Hence, the stopping potential of the material is 0.345 V.
(c) Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as: K = 1/2 mv2
Where,
m = Mass of an electron = 9.1 × 10−31 kg
v2 = 2K/m = 2 x 0.345 x 1.6 x 10-19 / 9.1 x 10-31 = 0.1104 x 1012
∴ v = 3.323 x 105 m/s = 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Stay updated with our latest educational content and study tips
Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 2: The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident ....
Comments
What is the energy of incident photons?
Thanks a lot