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Chapter 7: Q53E (page 282)

An electron is in the 3d state of a hydrogen atom. The most probable distance of the electron from the proton is $9 a_{o}$ . What is the probability that the electron would be found between $8 a_{o} a n d 10 a_{o}$ ?

Short Answer

Expert verified

The probability of finding the electron between $8 a_{o} a n d 10 a_{o} i s 0.212 .$

Step by step solution

01

Given data

The state of an electron 鈥� 3d.

02

Concept

The space around the proton in a hydrogen atom, in which the probability of finding the electron is at least $90 %$ , is known as the orbital.

03

Solution

The radial function is given as-

$R_{3, 2} (r) = \frac{1}{{(3 a_{o})}^{3 / 2}} \frac{(2 \sqrt{2 r^{2}})}{(27 \sqrt{5 a_{o}^{2}})} e^{- r / (3 a_{o})}$

The probability to find the electron between and

$P = {鈭�}_{8 a_{o}}^{10 a_{o}} {R_{3, 2}}^{2} (r) r^{2} d r = {鈭�}_{8 a_{o}}^{10 a_{o}} {(\frac{1}{{(3 a_{掳})}^{\frac{3}{2}}} \frac{2 \sqrt{2} r^{2}}{27 {\sqrt{5}}_{0}^{2}} e^{- r / 3 a_{掳}})}^{2} r^{2} d r = \frac{1}{{(3 a_{掳})}^{3}} \frac{8}{27^{2} 脳 5 a_{0}^{4}} {鈭�}_{8 a_{o}}^{10 a_{o}} r^{6} e^{- r / 3 a_{掳}} d r$

Let $\frac{r}{a_{o}} = x$ , then the equation becomes-

$P = \frac{8}{27^{3} 脳 5} {鈭�}_{8 a_{o}}^{10 a_{o}} X^{6} e^{2 x / 3} d r = 0.212$

This integral is found by the calculator.

Therefore, the probability of finding the electron between $8 a_{o} a n d 10 a_{o} i s 0.212 .$

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Most popular questions from this chapter

Can the transition $2 s 鈫� 1 s$ in the hydrogen atom occur by electric dipole radiation? The lifetime of the 2 s is known to be unusual. Is it unusually short or long?

In Table 7.5, the pattern that develops with increasing n suggests that the number of different sets of $(l, m_{l})$ values for a given energy level n is $n^{2}$ . Prove this mathematically by summing the allowed values of $m_{l}$ for a given $l$ over the allowed values of $l$ for a given n.

Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass $m_{1}$ , while $m_{2}$ is the mass of the orbiting negative charge. In positronium, an electron orbits a single positive charge, as in hydrogen, but one whose mass is the same as that of the electron -- a positron. Obtain numerical values of the ground state energy and 鈥淏ohr radius鈥� of positronium.

Here we Pursue the more rigorous approach to the claim that the property quantized according to m_l is L_z,

(a) Starting with a straightforward application of the chain rule,

$\frac{鈭�}{鈭� 蠁鈥�} = 鈥� \frac{鈭� x}{鈭� 蠁} / \frac{鈭�}{鈭� x} + \frac{鈭� y}{鈭� 蠁} \frac{鈭�}{鈭� y} + \frac{鈭� z}{鈭� 蠁} \frac{鈭�}{鈭� z}$

Use the transformations given in Table 7.2 to show that

$\frac{鈭�}{鈭� 蠁} 鈥� = 鈥� - y \frac{鈭�}{鈭� x} + x \frac{鈭�}{鈭� y}$

(b) Recall that L = r x p. From the z-component of this famous formula and the definition of operators for p_x and p_y, argue that the operator for L_z is $- i h \frac{鈭�}{鈭� 蠁} .$ .

(c) What now allows us to say that our azimuthal solution $e^{i m_{l} 蠁}$ has a well-defined z-component of angular momentum and that is value m_lh.

Doubly ionized lithium, $L i^{2 +}$ absorbs a photon and jumps from the ground state to its n=2level. What was the wavelength of the photon?

See all solutions

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