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Chapter 7: Q54E (page 282)

Using the functions given in Table 7.4, verify that for the more circular electron orbit in hydrogen $(i . e ., l = n - 1)$ , the radial probability is of the form
$P (r) 鈭� r^{2} n e^{- 2 r / n a_{o}}$
Show that the most probable radius is given by
$r_{m o s t p r o b a b l e} = n^{2} a_{o}$

Short Answer

Expert verified

The most probable radius is $r = n^{2} a_{0}$ .

Step by step solution

01

Given data

The radial function for $l = n - 1 i s,$

role="math" localid="1659182273339" $R_{n, n - 1} 鈭� r^{n - 1} e^{- r / (n a_{o})}$

02

Concept

The most probable radius is the radius of the orbit in which the probability of finding the electron is maximum.

The probability density is

$P (r) = r^{2} {R^{2}}_{n, n - 1}$

03

Solution

$P (r) = r^{2} {R^{2}}_{n, n - 1} = r^{2} (r^{n - 1} e^{- r / (n a_{o})}) = r^{2} r^{2 n - 2} e^{- r / (n a_{o})} = r^{2 n} e^{- r / (n a_{o})}$

To find the most probable location, take the derivative of the probability density,

$\frac{d P (r)}{d r} = \frac{d}{d r} r^{2 n} e^{- 2 r l (n a_{掳})} = 2 n r^{2 n - 1} e^{- 2 r l (n a_{掳})} + r^{2 n} (\frac{- 2}{n a_{掳}}) e^{- 2 r l (n a_{掳})} = e^{- 2 r l (n a_{掳})} r^{2 n - 1} (2 n + \frac{- 2 r}{n a_{掳}})$

For the most probable location, the derivative should be equal to zero.

role="math" localid="1659182748704" $= e^{- 2 r l (n a_{掳})} r^{2 n - 1} (2 n + \frac{- 2 r}{n a_{掳}})$

Hence,

$2 n + \frac{- 2 r}{n a_{掳}} = 0 r = n^{2} a_{掳}$

The most probable radius is $r = n^{2} a_{掳}$ .

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Most popular questions from this chapter

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. (a) What percentage error is introduced in the hydrogen ground-state energy by assuming that the proton is of infinite mass? (b) Deuterium is a form of hydrogen in which a neutron joins the proton in the nucleus, making the nucleus twice as massive. Taking nuclear mass into account, by what percent do the ground-state energies of hydrogen and deuterium differ?

Question: Consider an electron in the ground state of a hydrogen atom. (a) Calculate the expectation value of its potential energy. (b) What is the expectation value of its kinetic energy? (Hint: What is the expectation value of the total energy?)

Spectral lines are fuzzy due to two effects: Doppler broadening and the uncertainty principle. The relative variation in wavelength due to the first effect (see Exercise 2.57) is given by

$\frac{鈭� 位}{位} = \frac{\sqrt{3 k_{B} T / m}}{c}$

Where T is the temperature of the sample and m is the mass of the particles emitting the light. The variation due to the second effect (see Exercise 4.72) is given by

$\frac{鈭� 位}{位} = \frac{位}{4 蟺肠}$

Where, $鈭� t$ is the typical transition time

(a) Suppose the hydrogen in a star has a temperature of $5 脳 10^{4} K$ . Compare the broadening of these two effects for the first line in the Balmer series (i.e., $n_{i} = 3 鈫� n_{f} = 2$ ). Assume a transition time of 10^-8s. Which effect is more important?

(b) Under what condition(s) might the other effect predominate?

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?

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 $8 a_{o} a n d 10 a_{o}$ ?

See all solutions

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