# 30.3 Bohr’s theory of the hydrogen atom  (Page 7/14)

 Page 7 / 14

Explain how Bohr’s rule for the quantization of electron orbital angular momentum differs from the actual rule.

What is a hydrogen-like atom, and how are the energies and radii of its electron orbits related to those in hydrogen?

## Problems&Exercises

By calculating its wavelength, show that the first line in the Lyman series is UV radiation.

$\frac{1}{\lambda }=R\left(\frac{1}{{n}_{\text{f}}^{2}}-\frac{1}{{n}_{\text{i}}^{2}}\right)⇒\lambda =\frac{1}{R}\left[\frac{\left({n}_{\text{i}}\cdot {n}_{\text{f}}{\right)}^{2}}{{n}_{\text{i}}^{2}-{n}_{\text{f}}^{2}}\right];\phantom{\rule{0.25em}{0ex}}{n}_{\text{i}}=2,\phantom{\rule{0.25em}{0ex}}{n}_{\text{f}}=1,\phantom{\rule{0.25em}{0ex}}$ so that

$\lambda =\left(\frac{m}{1.097×{\text{10}}^{7}}\right)\left[\frac{\left(2×1{\right)}^{2}}{{2}^{2}-{1}^{2}}\right]=1\text{.}\text{22}×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{m}=\text{122 nm}$ , which is UV radiation.

Find the wavelength of the third line in the Lyman series, and identify the type of EM radiation.

Look up the values of the quantities in ${a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}{}^{}$ , and verify that the Bohr radius ${a}_{\text{B}}$ is $\text{0.529}×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{m}$ .

${a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kZq}}_{e}^{2}}=\frac{\left(\text{6.626}×{\text{10}}^{-\text{34}}\phantom{\rule{0.25em}{0ex}}\text{J·s}{\right)}^{2}}{{4\pi }^{2}\left(9.109×{\text{10}}^{-\text{31}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(8.988×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\text{·}{\text{m}}^{2}/{C}^{2}\right)\left(1\right)\left(1.602×{\text{10}}^{-\text{19}}\phantom{\rule{0.25em}{0ex}}\text{C}{\right)}^{2}}=\text{0.529}×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{m}$

Verify that the ground state energy ${E}_{0}$ is 13.6 eV by using ${E}_{0}=\frac{{2\pi }^{2}{q}_{e}^{4}{m}_{e}{k}^{2}}{{h}^{2}}\text{.}$

If a hydrogen atom has its electron in the $n=4$ state, how much energy in eV is needed to ionize it?

0.850 eV

A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is $n$ for a hydrogen atom if 0.850 eV of energy can ionize it?

Find the radius of a hydrogen atom in the $n=2$ state according to Bohr’s theory.

$\text{2.12}×{\text{10}}^{\text{–10}}\phantom{\rule{0.25em}{0ex}}\text{m}$

Show that $\left(13.6 eV\right)/\text{hc}=\text{1.097}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{m}=R$ (Rydberg’s constant), as discussed in the text.

What is the smallest-wavelength line in the Balmer series? Is it in the visible part of the spectrum?

365 nm

It is in the ultraviolet.

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.

Do the Balmer and Lyman series overlap? To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line.

No overlap

365 nm

122 nm

(a) Which line in the Balmer series is the first one in the UV part of the spectrum?

(b) How many Balmer series lines are in the visible part of the spectrum?

(c) How many are in the UV?

A wavelength of $4\text{.}\text{653 μm}$ is observed in a hydrogen spectrum for a transition that ends in the ${n}_{\text{f}}=5$ level. What was ${n}_{\text{i}}$ for the initial level of the electron?

7

A singly ionized helium ion has only one electron and is denoted ${\text{He}}^{+}$ . What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?

A beryllium ion with a single electron (denoted ${\text{Be}}^{3+}$ ) is in an excited state with radius the same as that of the ground state of hydrogen.

(a) What is $n$ for the ${\text{Be}}^{3+}$ ion?

(b) How much energy in eV is needed to ionize the ion from this excited state?

(a) 2

(b) 54.4 eV

Atoms can be ionized by thermal collisions, such as at the high temperatures found in the solar corona. One such ion is ${C}^{+5}$ , a carbon atom with only a single electron.

(a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen?

(b) What is the wavelength of the first line in this ion’s Paschen series?

(c) What type of EM radiation is this?

Verify Equations ${r}_{n}=\frac{{n}^{2}}{Z}{a}_{\text{B}}$ and ${a}_{B}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}=\text{0.529}×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{m}$ using the approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization.

$\frac{{\text{kZq}}_{e}^{2}}{{r}_{n}^{2}}=\frac{{m}_{e}{V}^{2}}{{r}_{n}}\text{,}$ so that ${r}_{n}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}{V}^{2}}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}}\frac{1}{{V}^{2}}\text{.}$ From the equation ${m}_{e}{\text{vr}}_{n}=n\frac{h}{2\pi }\text{,}$ we can substitute for the velocity, giving: ${r}_{n}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}}\cdot \frac{{4\pi }^{2}{m}_{e}^{2}{r}_{n}^{2}}{{n}^{2}{h}^{2}}$ so that ${r}_{n}=\frac{{n}^{2}}{Z}\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}=\frac{{n}^{2}}{Z}{a}_{\text{B}},$ where ${a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}$ .

The wavelength of the four Balmer series lines for hydrogen are found to be 410.3, 434.2, 486.3, and 656.5 nm. What average percentage difference is found between these wavelength numbers and those predicted by $\frac{1}{\lambda }=R\left(\frac{1}{{n}_{\text{f}}^{2}}-\frac{1}{{n}_{\text{i}}^{2}}\right)$ ? It is amazing how well a simple formula (disconnected originally from theory) could duplicate this phenomenon.

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Matthew, photons ARE light. there is no such thing as a photon that isn't moving. in fact the speed they move at is called C (for constant) in physics. through a vacuum they always travel at this speed no matter what. they can not slow down; except in another medium.
The reason why a photon can go at this speed is BECAUSE it had no mass. nothing can go this speed or faster because it needs to have no mass or negative mass. that's why it's called the constant.
when a photon hits something that is opaque, this is the only way to "stop"it. it isn't merely stopped but absorbed and turned into heat energy, then the remaining energy is reflected in different wavelengths. that reflection is what we call color. the darker something is, the less photons are ther
e. complete blackness is the absolute absence of photons altogether. I believe what you're referring to is not speed, but wavelength, which is indirectly proportional to the amount of energy a particular photon is made up of.
in order for a photon to have zero wavelength, it must (at least theoretically) have infinite energy.
about mass: you may have photons confused with electrons. elections have a mass so small that people say they are without mass, but they do. it is called electron mass or Me-.
you may also be getting electrons and photons confused because of the cherenkov effect. that is what happens when a particle travels faster than light IN THAT PARTICULAR MEDIUM. I emphasize that because no other particle besides photons can go the speed of c.
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is very close to c, which is the universal speed limit. I'd you did go faster than c, time would go backwards and you would have infinite theoretical mass and probably spagghettify, like with a black hole.
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