By Ahmad A. Kamal

ISBN-10: 3642043321

ISBN-13: 9783642043321

This ebook essentially caters to the wishes of undergraduates and graduates physics scholars within the sector of recent physics, specifically particle and nuclear physics. Lecturers/tutors might use it as a source booklet. The contents of the booklet are in line with the syllabi at the moment utilized in the undergraduate classes in united states, U.K., and different nations. The publication is split into 10 chapters, every one bankruptcy starting with a short yet enough precis and worthy formulation, tables and line diagrams by way of numerous regular difficulties precious for assignments and checks. targeted recommendations are supplied on the finish of every chapter.

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**Extra info for 1000 Solved Problems in Modern Physics**

**Example text**

R −3 = (x iˆ + y ˆj + z k) ∂x ∂y ∂z ˆ . − 3 . (2x iˆ + 2y ˆj + 2z k)(x ˆ 2 + y 2 + z 2 )−5/2 = (x iˆ + y ˆj + z k) 2 = −3(x 2 + y 2 + z 2 )(x 2 + y 2 + z 2 )− 2 = −3r −3 Thus ∇ . 4 By problem ∇ × A = 0 and ∇ × B = 0, it follows that B . (∇ × A) = 0 A. (∇ × B) = 0 Subtracting, B . (∇ × A) − A . (∇ × B) = 0 Now ∇ . ( A × B) = B . (∇ × A) − A . (∇ × B) Therefore ∇ . ( A × B) = 0, so that ( A × B) is solenoidal. 5 (a) Curl {r f (r )} = ∇ × {r f (r )} = ∇ × {x f (r )iˆ + y f (r ) ˆj + z f (r )k} ˆj iˆ kˆ ∂ ∂ ∂ ∂f ∂f = z −y = ∂x ∂y ∂z ∂y ∂z x f (r ) y f (r ) z f (r ) But ∂f ∂x = ∂f ∂r ∂r ∂x ∂f = yrf ∂y Similarly respect to r .

2 0 r 2(m+n)−1 e−r dr 2 (4) 0 In (4), the first integral is identified as B(m, n) and the second one as Γ (m + n). 27 Let H be the hermitian matrix with characteristic roots λi . 3 Solutions 45 H X i = λi X i Now X¯ ι H X i = X¯ ι λi X i = λi X¯ ι X i (1) (2) is real and non-zero. 28 The characteristic equation is given by 1 − λ −1 1 0 3 − λ −1 = 0 0 0 2−λ (1 − λ)(3 − λ)(2 − λ) + 0 + 0 = 0 or λ3 − 6λ2 + 11λ − 6 = 0 (characteristic equation) (1) (2) The eigen values are λ1 = 1, λ2 = 3, and λ3 = 2.

69 Find the solution to the differential equation: 3 dy + y = x +2 dx x +2 which satisfies y = 2 when x = −1, express your answer in the form y = f (x). 70 (a) Find the solution to the differential equation: dy d2 y −4 + 4y = 8x 2 − 4x − 4 2 dx dx = 0 when x = 0. 72 Consider the chain decay in radioactivity A → B → C, where λ A and λ B are the disintegration constants. The equations for the radioactive decays are: dN B (t) dN A (t) = −λ A N A (t), and = −λ2 N B (t) + λ A N A (t) dt dt where N A (t) and N B (t) are the number of atoms of A and B at time t, with the initial conditions N A (0) = N A0 ; N B (0) = 0.

### 1000 Solved Problems in Modern Physics by Ahmad A. Kamal

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