Is anyone able to provide a worked solution for this question? It\'s from Fundam

Question

Is anyone able to provide a worked solution for this question? It's from Fundamentals of Physics 10th Edition Extended. Chapter 39 problem 66. Thanks.

In atoms there is a finite, though very small, probability that, at some instant, an orbital electron will actually be found inside the nucleus. In fact, some unstable nuclei use this occasional appearance of the electron to decay by electron capture. Assuming that the proton itself is a sphere of radius 1.1 × 10-15 m and that the wave function of the hydrogen atom's electron holds all the way to the proton's center, use the groundstate wave function to calculate the probability that the hydrogen atom's electron is inside its nucleus.

Explanation / Answer

Radius of hydrogen atom nucleus a0 = 1.1*10^-15 m

the ground state wave function of hydrogen atom

Y100 = A (1/ a0^3)^1/2 *2 *e^-r/a0

probability of finding the electron inside the nucleus is

<Y100|Y100> = 1

A^2 *4*(1/ a0^3) integral of e^-2r/a0*r^2 sin theta d theta dphi dr = 1

A^2 *4 * (1/pi a0^3) 4*pi integral of e^-2r/a0 r^2 dr =1

A^2 *4*pi/a0^3 *(a0/2)^3 integral of e^-x *x^2 dx = 1 x= 2r/a0

A^2 * 4pi * 1/2 * 2! = 1

A^2 = 1 / 4pi

A^2 = 0.079

the probability of finding the electron inside the nucleus is 0.079 is 7.9%

Y 100 = (1/pi a0^3)^1/2 *e^-r/a0

we have to find the average value < r > for electron to be located

< r > = < Y100 | r | Y100>

= (1/pi a0^3) integral of r e^-2r/a0 r^2 sin theta dr dphi dtheta

= 1/pia0^3 * 4 pi integral of e^-2r/a0 r^3 dr

= 4/a0^3 * ( a0/2)4 integral of e^-x *x^3 dx x = 2r/a0

= a0/4 x 3! integral of e^-x *x^n = n! ( gamma function)

< r > = 3/2 a0

< r^2> = <Y100 | r^2 | Y100 >

= (1/pi a0^3) integral of r^2* e^-2r/a0 *r^2 sin theta dtheta dphi dr

= (1/pi a0^3) 4 pi integral of e^-2r/a0* r^4 dr

= 4/a0^3 ( a0/2)^5 integral of e^-x* x^5 dx x = 2r/a0

= a0^2/8 * 4!

< r2 > = 3a0^2

Uncertainity in position to finding the particle inside the nucleus

delta r = sqrt( < r^2> - < r >^2)

= sqrt ( 3a02 - (3a0/2 )^2 )

= sqrt ( 3 a0^2 - 9a0^2/4 )

delta r = sqrt (3 a0 /2)

delta r = sqrt (3 *1.1*10^-15 / 2)

delta r = 1.284*10^-15 m

uncetainity in postion to finding the particle inside the nucleus 1.284*10^-15 m which less than the radius of nucleus.

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