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 1: . 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.
Potential of the electrons, V = 30 kV = 3 × 104 V
Hence, energy of the electrons, E = 3 × 104 eV
Where,
e = Charge on an electron = 1.6 × 10−19 C
(a)Maximum frequency produced by the X-rays = ν
The energy of the electrons is given by the relation:
E = hν
Where,
h = Planck’s constant = 6.626 × 10−34 Js
So v = E / h
= 1.6 x 10-19 x 3 x 104 / 6.626 x 10-34 = 7.24 x 1018 Hz
Hence, the maximum frequency of X-rays produced is 7.24 x 1018 Hz
(b)The minimum wavelength produced by the X-rays is given as:
λ = c / v
= 3 x 108 / 7.24 x 1018 = 4.14 x 10-11 m = 0.0414 nm
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.
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Read MoreWelcome 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 1: Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.....
Comments
instead of 104 read it 10^4
Answer should be hindi ok
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