Home Page Textbooks Physics Page 893: Practice Problems

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 893: Practice Problems

Exercise 9

Solution 1

Solution 2

Step 1

1 of 2

To be able to know the wavelength of the red spectral line (corresponds to the jump from n = 3 to n = 2), we use the following formula:

$$
begin{align}
Delta E &= E_2 – E_3 \
&= 13.6 text{ eV} times left(dfrac{1}{2^2} – dfrac{1}{3^3}right) \
Delta E &= dfrac{hc}{lambda}
end{align}
$$

Using these equations, we can derive an equation which can solve for $lambda$, such that:

$$
begin{align*}
lambda &= dfrac{hc}{13.6 text{ eV} left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{eV}}right) left(dfrac{1}{2^2} – dfrac{1}{3^3}right)} \
&= dfrac{(6.626 times 10^{-34} text{ Js})(3.0 times 10^8 text{m/s})}{13.6 text{ eV} left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{ eV}}right) left(dfrac{1}{2^2} – dfrac{1}{3^3}right)} \
&= 657 text{ nm}
end{align*}
$$

Therefore, the wavelength of the red spectral line is $boxed{text{657 nm}}$

Result

2 of 2

$$
textbf{657 nm}
$$

Step 1

1 of 2

The energy of this line is exactly the difference between the energies of $3$rd and $2$nd Bohr’s orbits. Knowing that this energy is of the form $hnu= hfrac{c}{lambda}$ we can write

$$
frac{hc}{lambda} =E_3-E_2 = 13.6text{ eV}left(frac{1}{2^2}-frac{1}{3^2}right).
$$
This yields
$$
lambda = frac{hc}{13.6text{ eV}left(frac{1}{2^2}-frac{1}{3^2}right)} = frac{hc}{13.6times1.6times10^{-19}text{ J}left(frac{1}{2^2}-frac{1}{3^2}right)} = 657text{ nm}.
$$

Result

2 of 2

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Exercise 10

Solution 1

Solution 2

Step 1

1 of 2

To be able to know the frequency of light (corresponds to the quantum jump from n = 5 to n = 4), we use the following formula:

$$
begin{align}
Delta E &= E_4 – E_5 \
&= 13.6 text{ eV} times left(dfrac{1}{4^2} – dfrac{1}{5^3}right) \
Delta E &= hv
end{align}
$$

Using these equations, we can derive an equation which can solve for $v$, such that:

$$
begin{align*}
v &= dfrac{13.6 text{ eV} left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{eV}}right) left(dfrac{1}{4^2} – dfrac{1}{5^3}right)}{h} \
&= 7.4 times 10^{13} text{ Hz}
end{align*}
$$

Therefore, the frequency is $boxed{7.4 times 10^{13} text{ Hz}}$

Result

2 of 2

$$
7.4 times 10^{13} text{ Hz}
$$

Step 1

1 of 2

Knowing that this energy is of the form of $hnu$ ($nu$ is the required frequency) and that it is exactly equal to the difference between energies of the 5th and 4th Bohr’s orbit we can write

$$
hnu = E_5-E_4=13.6text{ eV}left(frac{1}{4^2}-frac{1}{5^2}right)
$$
and thus obtain

$$
nu=frac{13.6text{ eV}left(frac{1}{4^2}-frac{1}{5^2}right)}{h} =frac{13.6times 1.6times10^{-19}text{ J}left(frac{1}{4^2}-frac{1}{5^2}right)}{h}=7.4times10^{13}text{ Hz}.
$$

Result

2 of 2

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Exercise 11

Step 1

1 of 2

The higher the energy of the emitted light the lower it’s wavelength which is seen from $E=hc/lambda$. Since the energy difference between the 7th and the 2nd orbit is greater then the difference between the 4th and the 2nd (because 7th orbit has higher energy than 4th) the jump $7to2$ carries more energy than $4to2$. This means that the emitted light in the 1st case carries more energy and thus has lower wavelength.

Result

2 of 2

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