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Chapter 41: Q. 3 (page 1206)

What is the maximum possible angular momentum $L$ (as a multiple of h) of a hydrogen atom with energy $- 0.544 e V$ ?

Short Answer

Expert verified

The maximum possible angular momentum $L$ is $\sqrt{20 h}$ .

Step by step solution

01

Given Information

We need to find the maximum possible angular momentum $L$ .

02

Simplify

Finding the maximum angular momentum $L$ of a hydrogen atom.

Consider:

$E = - 0.544 e V$

know that the hydrogen atom's energy is given as:

$E_{n} = - \frac{13.6 e V}{n^{2}}$

Using formula, find the value of $n$ :

$E_{n} = - \frac{13.6 e V}{n^{2}} - 0.544 e V = - \frac{13.6 e V}{n^{2}} (- 0.544 e V) 脳 n^{2} = - 13.6 e V n^{2} = \frac{- 13.6 e V}{- 0.544 e V} n^{2} = 25 n = \sqrt{25} n = 5$

Now, finding the maximum angular momentum $L$ .

Angular momentum $L$ is given as:

$L = \sqrt{l (l + 1) h}$

Then,

$L = \sqrt{l (l + 1) h} L = \sqrt{4 (4 + 1) h} L = \sqrt{4 (5) h} L = \sqrt{20 h}$

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Most popular questions from this chapter

Draw a series of pictures, similar to the ground states of $K, S c, C o$ and $G e$ .

$1.0 脳 10^{6}$ sodium atoms are excited to the $3 p$ state at $t = 0$ s. How many of these atoms remain in the $3 p$ state at

(a) $t = 10 n s$

(b) $t = 30 n s$ , and

(c) $t = 100 n s$ ?

Identify the element for each of these electron configurations. Then determine whether this configuration is the ground state or an excited state.

a. $1 s^{2} 2 s^{2} 2 p^{5} 3 s$

b. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{2}$

What is the angular momentum of a hydrogen atom in

(a) a $6 s$ state and

(b) a $4 f$ state? Give your answers as a multiple of U.

Find the distance, in terms of $a_{B}$ , between the two peaks in the radial probability density of the $2 s$ state of hydrogen.

Hint: This problem requires a numerical solution.

See all solutions

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