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Academic session 2026-27, Chapter 11 in the current curriculum.
Welcome 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 12:
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
. 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.Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Potential difference, V = 56 V
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:
1/2 mv2 = eV
v2 = 2eV / m

The momentum of each accelerated electron is given as:
p = mv
= 9.1 × 10−31 × 4.44 × 106
= 4.04 × 10−24 kg m s−1
Therefore, the momentum of each electron is 4.04 × 10−24 kg m s−1 .
(b) De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
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:
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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 12: Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through....
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