13 meters Physics 107 Problem 4.7 O

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We can use equation 4.2, modified to account for the +1 charge on the proton.

r 0 Z e. ² 4. επ. 0.K

Define the constants and variables:

Z 79 e 1.60 10. ¹⁹ ε0 8.85 10. ¹²

in MKS units K 1 10. ⁶. 1.602 10. ¹⁹ this converts 1 MeV to MKS

r 0 Z e. ² 4. επ. 0.K

r 0 1.135 10= . ¹³ meters

Physics 107 Problem 4.7 O. A. Pringle

The answer to this question is best found in section 4.2. Beiser gives the kinetic and potential energies of the electron in page 124.

The kinetic energy, mv^2/2, is always positive, and the electron is, indeed, in motion.

The potential energy, however, results from the Coulomb attraction of the proton and electron. This negative energy is larger that the kinetic energy.

The total energy E=K+V is therefore negative, indicating the electron is bound, but the kinetic energy is positive, so the electron can be in motion.

We want to begin with L n hbar. and arrive at r n

n². εh². 0 π.m.e² Begin:

L n hbar. m v. .r n r n n h.

2.π.m.v Plug in v from eq. 4.4

r n² n².h² 4.π².m² e²

4. επ. 0.m.r n .

r n² r n

n². εh². 0

π.m.e² r n

n². εh². 0

π.m.e² The desired result.

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Physics 107 Problem 4.9 O. A. Pringle This solution is algebraic. Paper and pencil might be easier than Mathcad.

Equation 4.4 gives the velocity of an orbiting electron.

v e

4. επ. 0.m.r

All we need to do is substitute in the Bohr radius of the orbit.

r n n².a 0

v n e

4. επ. 0.m.r n

v n e

4. επ. 0.m.n².a 0

v n e

n. 4. επ. 0.m.a 0

Sample calculation with numbers: e 1.6 10. ¹⁹ ε0 8.85 10. ¹² m 9.11 10. ³¹ a 0 5.29 10. ¹¹

v n( ) e

n. 4. επ. 0.m.a 0 v 1( )=2.186 10. ⁶ v 5( )=4.371 10. ⁵

First, I will find the earth's wavelength. Then I will use the Bohr model equation relating wavelength and orbital radius to find the earth's quantum number.

h 6.63 10. ³⁴ m earth 6 10. ²⁴ v earth 3 10. ⁴ r orbit 1.5 10. ¹¹ λearth

h m earth v earth.

λearth=3.683 10. ⁶³ I changed the zero tolerance to 100 to see something besides zero here.

The earth can "fit into" an orbit only if nλ=2πr_n.

n earth

2.π.r orbit λearth

n earth 2.559 10= . ⁷⁴

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m electron 9.11 10. ³¹

a 0 5.29 10. ¹¹ Note the inside front cover of Beiser has a typo in its value for a₀.

According to the uncertainty principle, the uncertainty in the momentum of an electron in a ground-state Bohr atom (taking a₀ as an estimate of the size of the region to which the electron is confined) is at least

δp h

4.π.a 0

δp =9.974 10. ²⁵ In MKS units, a small number.

We can calculate the momentum of this electron in several ways. We know its speed, so we can get its momentum from p=mv.

v electron 2.2 10. ⁶

p electron m electron v electron. p electron 2.004 10= . ²⁴

p electron δp

=2.01 The actual momentum is a factor of 2 bigger than the uncertainty.

The equation giving the wavelength of a photon emitted by an electronic transition in hydrogen is

λ E 1

c h. 1 n f²

1 n i² .

1

(equation 4.18, page 134)

In this problem, the electron is initially unbound; i.e.

n i ∞ or 1

n i 0

The electron goes from the unbound state to the ground (n=1) state, so

I like to manipulate Mathcad equations first to get the equation I will finally use. I like to do this before I set up the rest of the problem, so I can see what I need to know.

λ E 1

c h. 1 n f² .

1

Now define the parameters and plug them in.

h 6.63 10. ³⁴ c 3 10. ⁸ n f 1 E 1 13.6 1.60. .10 ¹⁹

λ E 1

c h. 1 n f² .

1

λ=9.141 10. ⁸ or about 91.4 nm

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Physics 107 Problem 4.19 O. A. Pringle I "borrowed" the equation from the solution to problem 4.20, which I did before 4.19.

We are given: n i 10 n f 1 R 1.097 10. ⁷

λ R 1

n f² 1 n i²

. ¹

λ=9.208 10. ⁸ or 92.1 nm; this is a short wavelength, so it corresponds to high energy, or ultraviolet

Equation 4-18 gives λ: 1

λ E 1 c h.

1 n f²

1 n i²

. λ E 1

c h. 1 n f²

1 n i² .

1

λ R 1

n f² 1 n i²

. ¹

where R 1.097 10. ⁷m^-1 We are given: n i 6 n f 3

λ R 1

n f² 1 n i²

. ¹

λ=1.094 10. ⁶ or 1094 nm

According to figure 2.2 on page 51, this wavelength would correspond to relatively high-energy infrared, a little below the visible (which starts at about 400 nm).

As usual, the first step is to figure out what the problem is really asking.

It's really just asking you to calculate the energy of the photon corresponding the the "first" line of the Balmer series.

What does Beiser mean by "first line?" If you look at fig. 4-16, you will see that the series limit is approached as n-->∞. Thus, there is no "last line" of the Balmer series, and the "first line" must correspond to the n=3 to n=2 transition.

There are several ways to solve this problem. You could use eq. 4.18 to get λ of the photon, and use E=hc/λ to get the photon (and therefore electron) energy. Or you could use eq. 4.17 to get the frequency and use E=hf. Or you could use the un-numbered equation between equations 4.16 and 4.17. That's what I will do. That's handy because I don't have to go through any energy conversions.

E 1 13.6

∆E n i n f, E 1 1 n f²

1 n i²

. ∆E 3 2( , )=1.889 volts

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As usual, the first step is to figure out what the problem is really asking.

The energy of an electron in hydrogen is given by E n( ) E 1

n² where E 1 13.6 eV An electron in the n=2 state has E 2( )= 3.4 To remove it from the hydrogen atom requires 3.4 eV:

∆E E(∞) E 2( )

∆E E(∞) E 2( ) I did this on purpose. Mathcad thinks 10³⁰⁷ is infinity. I tried typing in control-z (the way to get "infinity". Mathcad says "overflow" because it can't handle 1/(infinity)². To get around this, just realize E(∞)=0.

∆E 0 E 2( )

∆E=3.4 eV

The pertinent equation is 1

λ

E 1 h c.

1 n f²

1 n i² .

Because E₁ is the reduction in energy of the electron upon binding, the ionization energy is actually the negative of E₁:

1 λ

E ionization h c.

1 n f²

1 n i² .

The Lyman series corresponds to transitions from n of 2 or greater down to n=1. The longest wavelength would correspond to a transition from n=2 to n=1, the smallest-energy transition in the series.

Solving for E_ionization gives

E ionization λ,n f,n i h c. λ

1 1 n f²

1 n i² .

E ionization 121.5 10. ⁹,1,2 =2.183 10. ¹⁸ Does this answer make any sense?

E in_eV E ionization 121.5 10. ⁹,1,2 e

E in_eV 13.6= eV yes, the answer does make sense

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Physics 107 Problem 4.25 O. A. Pringle Part (a)

I suppose to be consistent with the other constants, I could round this to 1.10x10⁷.

1

λ R 1

n f² 1 n i²

. R 1.097 10. ⁷

The final state is the ground state, n_f=1. We assume R and λ are known, and solve for n_i.

1 R.λ

1 1

1 n i² 1

n i²

1 1

R.λ

n i² 1 1

R.λ

1

Notice this is a "live" equation, n_i as a function of λ. n i( )λ 1 1

R.λ

1

Part (b) n i 102.55 10. ⁹ =3

h 6.63 10. ³⁴ c 3 10. ⁸ λ 694 10. ⁹ The energy of a 694 nm photon is

E h c.

λ E=2.866 10. ¹⁹Joules

In each pulse of light, there are

n 1

E

photons, because 1 joule per pulse divided by joules per photon gives photons per pulse.

n=3.489 10. ¹⁸ photons

There must be at least that many chromium 3+ ions in the crystal.

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This sounds tricky, but it is just asking: what is the frequency of a photon whose energy is equal to the energy given up by when a molecule of steam turns to water?

This is really just a unit conversion problem.

M 18.02 grams per mole N 6.02 10. ²³molecules per mole

Let m be the number of kilograms per molecule.

m M 10. ³ N

Is this reasonable? H₂O ought to be about 18 times the mass of hydrogen:

m=2.993 10. ²⁶

18 1.67. .10²⁷=3.006 10. ²⁶so our m is reasonable.

L is the heat of vaporization:

L 2260 10. ³ joules/kilogram E m L.

I really only need to define this once per document, don't I.

E=6.765 10. ²⁰ joules h 6.63 10. ³⁴

f E

h

According to figure 2.2 on page 47, this would be high-frequency infrared light, almost visible.

f=1.02 10. ¹⁴