Home Page Textbooks Physics Page 890: Practice Problems

Physics

1st Edition

Walker

ISBN: 9780133256925

Textbook solutions

All Solutions

Page 890: Practice Problems

Exercise 5

Solution 1

Solution 2

Step 1

1 of 2

To be able to compute for the n-orbit Bohr radius ($r_n$), we use the formula:

$$
begin{gather}
r_{n} = (5.29 times 10^{-11} text{ m}) times n^2
end{gather}
$$

Using Eq (1), we can solve for n, such that:

$$
begin{align*}
n &= sqrt{dfrac{1.90 times 10^{-9} text{ m}}{5.29 times 10^{-11} text{ m}}} \
&= 6
end{align*}
$$

Hence, the Bohr orbit that has a radius of $1.90 times 10^{-9}$ m is $boxed{text{n = 6}}$

Result

2 of 2

n = 6

Step 1

1 of 2

We know that each Bohr’s radius is of the form of
$$
r_n = (5.29times10^{-11}text{ m})n^2.
$$

This yields

$$
n=sqrt{frac{r_n}{5.29times10^{-11}}}=sqrt{36}=6
$$

so this is sixth Bohr’s orbit of radius $r_n=r_6.$

Result

2 of 2

Click here for the solution.

Exercise 6

Solution 1

Solution 2

Step 1

1 of 2

$textbf{Given:}$

$r_{n1} = 9.5 times 10^{-10}$ m

$r_{n2} = 1.7 times 10^{-9}$ m

$textbf{Find:}$ $r_n$ such that:

$$
begin{gather}
r_{n1} < r_n < r_{n2}
end{gather}
$$

To be able to compute for the n-orbit Bohr radius ($r_n$), we use the formula:

$$
begin{gather}
r_{n} = (5.29 times 10^{-11} text{ m}) times n^2
end{gather}
$$

Deriving Eq (2), we can solve for $n_1$ and $n_2$, such that:

$$
begin{align*}
n_1 &= sqrt{dfrac{9.5 times 10^{-10} text{ m}}{5.29 times 10^{-11} text{ m}}} \
&= 4 approx 4 \
n_2 &= sqrt{dfrac{1.7 times 10^{-9} text{ m}}{5.29 times 10^{-11} text{ m}}} \
&= 5.67 approx 6
end{align*}
$$

Applying the relationship (1), we have:
$4 < n < 6$

Hence, the Bohr orbit that lies in this range is $boxed{text{n = 5}}$

Result

2 of 2

$$
textbf{n = 5}
$$

Step 1

1 of 2

We know that this radius has to be have the form of $r_n=(5.29times10^{-11}text{ m})n^2$.We know that

$$
9.5times 10^{-10}text{ m}<r_n<1.7times 10^{-9}text{ m}.
$$

Dividing this by
$$
5.29times 10^{-11}text{ m}
$$
we get

$$
18<n^2<32.
$$
The only natural number satisfying this is $n=5$ yielding

$$
r_5=(5.29times10^{-11}text{ m})times 25 = 1.32times 10^{-9}text{ m}.
$$

Result

2 of 2

Click here for the solution.

unlock

All Solutions

Page 890: Practice Problems

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards