Chapter 39: Q.48 (page 1140)

The probability density of finding a particle somewhere along the $x - a x i s i s 0 f o r x 61 m m . A t x = 1 m m,$ the probability density is $c$ . For $x U 1 m m$ , the probability density decreases by a factor of $8$ each time the distance from the origin is doubled. What is the probability that the particle will be found in the interval $2 m m . . . x . . . 4 m m ?$

Short Answer

Expert verified

$p (x) i s p r o b a b i l i t y d e n s i t y$

Step by step solution

Probability of finding a particle in certain interval δx is given by

probability to finding a particle in $a \leq x \leq b = \int_{a}^{b} p (x) d x$

Calculate the value C by using normalization condition,

$\int_{- \infty}^{+ \infty} p (x) d x = 1 T h e i n t e g r a l c a n b e s u b d i v i d e d i n t e r m s o f t w o i n t e g r a l, \int_{- \infty}^{1} p (x) d x + \int_{1}^{\infty} p (x) d x = 1 N o w, S u b s t i t u t e v a l u e o f p (x) \int_{- \infty}^{1} (0) d x + \int_{1}^{\infty} (\frac{c}{x^{3}}) d x = 1 {[\frac{c}{- 2 x^{2}}]}_{1}^{\infty} = 1 c = 2$

now, calculate probability of finding particle in interval 2mm $\leq x \leq 4 m m$ as follows,

$p r o b a b i l i t y (2 m m \leq x \leq 4 m m) = \int_{2}^{4} (\frac{c}{x^{3}}) d x = {[\frac{c}{- 2 x^{2}}]}_{2}^{4} = [\frac{c}{- 2 {(4)}^{2}} - \frac{c}{- 2 {(2)}^{2}}] = \frac{c (3)}{32} N o w, s u b s t i t u t e, 2 f o r c i n t h e a b o v e r e s u l t p r o b a b i l i t y (2 m m \leq x \leq 4 m m) = \frac{(2) (3)}{32} = \frac{3}{16} H e n c e, p r o b a b i l i t y o f f i n d i n g p a r t i c l e i n i n t e r v a l 2 m m \leq x \leq 4 m m i s \frac{3}{16}$