(1)Physics 107 Problem 2.5 O

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Physics 107 Problem 2.5 O. A. Pringle

c 3 10. ⁸ h 6.63 10. ³⁴ λ 700 10. ⁹

f c

λ E h f.

E=2.841 10. ¹⁹ Joules Note I had to set the zero tolerance here.

e 1.6 10. ¹⁹ eV <-> joules conversion factor E eV E

e E eV 1.776= eV

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E h f. h 6.63 10. ³⁴ c 3 10. ⁸

f E

h

E eV 100 10. ⁶eV e 1.6 10. ¹⁹ E E eV e.

f E

h

f=2.413 10. ²²Hz c f.λ so λ c

f

λ c

f

λ=1.243 10. ¹⁴meters in nanometers, λnm λ.10⁹

λnm=1.243 10. ⁵

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h 6.63 10. ³⁴ f 880 10. ³

photon energy E h f.

Power is energy per time. Photons per second is energy per time divided by energy per photon.

P 1 10. ³

n P

E

n=1.714 10. ³⁰ photons per second

Note that as long as I keep everything in mks units, the units automatically work out OK.

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h 6.63 10. ³⁴ E 10¹⁸

λ 600 10. ⁹ c 3 10. ⁸

E photon h c.

λ E photon 3.315 10= . ¹⁹

N E

E photon

N=3.017 The eye can detect 3 of these photons.

Note again that units work out OK if we stick with the mks system.

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(a) How many photons fall per second on each square meter of the earth's surface directly facing the sun?

h 6.63 10. ³⁴ f photon 5 10. ¹⁴

on each square meter: P 1.4 10. ³

as in problem 2.5, number per second is power divided by energy n P h f photon. n=4.22 10. ²¹ photons/s

(b) What is the power output of the sun, and how many photons per second does it emit?

The power per area at a radius of 1.5*10^11 m is given as 1.4x10^3 W/m^2. To answer the first part, just multiply the power per area by the area of a sphere of the given radius.

r 1.5 10. ¹¹ A 4.π.r²

P sun P A.

P sun 3.96 10= . ²⁶watts

N P sun

h f photon.

N=1.19 10. ⁴⁵ photons per second emitted by the sun

(c) How many photons per cubic meter are there near the earth? c 3 10. ⁸will need this in a minute This is mainly a unit conversion problem. Look at the units.

number volume

number s

s volume

= .

number volume

number s

s m

. 1

m²

= .

density number s

1 speed

. 1

area

= .

ρ n 1

c

. 1

1² .

ρ=1.41 10. ¹³ photons per cubic meter

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Physics 107 Problem 2.10 O. A. Pringle t 20 10. ³ P 0.5 λ 632 10. ⁹

h 6.63 10. ³⁴ c 3 10. ⁸

The energy of a pulse is the power times the time.

E pulse P t.

The number of photons in the pulse is the energy of the pulse divided by the energy of a photon.

N E pulse h c.

λ N=3.18 10. ¹⁶

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The equation to use is K max h f. h f 0.

energies will come out in eV if I use h in these units

h 4.14 10. ¹⁵ c 3 10. ⁸ λ0 230 10. ⁹

f 0 c λ0 K max 1.5

Solve for f f K max h f 0. h

f=1.667 10. ¹⁵here's the frequency; not necessary, just wanted to look at it

λ c

f λ=1.8 10. ⁷

This is a wavelength of 180 nm.

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h 4.14 10. ¹⁵ will give K in eV We are given the following:

f 0 1.1 10. ¹⁵ f 1.5 10. ¹⁵ Simply plug these into equation 2.1

K max h f. h f 0. K max 1.656= eV

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Physics 107 Problem 2.13 O. A. Pringle The equation we will use is

K max h f. h f 0.

where h 4.14 10. ¹⁵ will give us energies in eV

Because E=hf and f=c/λ, longer wavelengths of light have lower energies. This problem is equivalent to asking

"what is the minimum frequency of light that will cause photoelectrons to be emitted from sodium." That minimum frequency is just the threshold frequency for sodium, which can be found from the work function.

φ 2.3 eV, from table 2.1 c 3 10. ⁸

This is just the minimun energy needed to produce a photoelectron, so φ h f 0.

or f 0 φ

h f 0 5.556 10= . ¹⁴ Hz The maximum wavelength is just, using

λ0 c f 0

λ0=5.4 10. ⁷meters, or 540 nm (to get Beiser's answer, use c=2.998*10^8 and h=4.136*10^-15) What will the maximum kinetic energy of the photoelectrons be if 200-nm light falls on a sodium surface?

This is just like problem 2.10, except we are given wavelengths instead of frequencies.

λ0=5.4 10. ⁷ calculated above λ 200 10. ⁹

K max h c. λ

h c. λ0

K max 3.91= eV

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This sounds tricky but really isn't. Light incident on the ball will cause photoelectrons to be emitted.

The ball will acquire a positive charge and therefore an electrical potential as the electrons are emitted.

When work function plus the potential of the ball equals the energy of the incident light, no more electrons will be emitted.

For silver φ 4.7 eV

Planck's constant h 4.14 10. ¹⁵ eV*s frequency of incident light:

c 3 10. ⁸

λ 200 10. ⁹ f c

λ f=1.5 10. ¹⁵ Hz In words: incident energy=V+φ

V h f. φ V=1.51Volts

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Physics 107 Problem 2.15 O. A. Pringle The 1.5 mW (milliwatts) gives the energy per unit time in the incident light beam.

P 1.5 10. ³ joules/s 400-nm tells us the photon energy.

h 6.63 10. ³⁴I'll work in mks units here.

c 3 10. ⁸ λ 400 10. ⁹ E photon h c.

λ

E photon 4.972 10= . ¹⁹joules per photon

Dividing the power (energy per time) by the energy of a photon (energy per photon) gives us the number of photons per second incident on the cell.

N P

E photon

N=3.017 10. ¹⁵ photons per second

Only 0.1 percent of these photons produce photoelectrons, so the number n producing photoelectrons is

n N

1000

n=3.017 10. ¹²electrons produced per second

This n is actually the current, but we should express it in the more familiar units of coulombs e 1.60 10. ¹⁹ coulombs per electron

I n e.

I=4.827 10. ⁷ coulombs per second, or amps

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a) Find the extinction voltage, that is, the retarding voltage at which the photoelectron current disappears.

The extinction voltage occurs when the retarding voltage plus the work function equal the photoelectron energy.

φ V ext h f.

h 4.14 10. ¹⁵ c 3 10. ⁸

φ 2.50

λ 400 10. ⁹ f c V ext h f. φ λ

V ext 0.605= electron volts b) Find the speed of the fastest photoelectrons.

The most energetic electrons will appear when V.ext=0, i.e., K max h f. φ

K max 0.605= electron volts; yes, the same number as in part a

Experience should tell us that 0.6 eV is nonrelativistic; we will do a nonrelativistic calculation, and if the speed is too great, go back and do a relativistic calculation.

K max 1

2.m electron.v max² e 1.6 10. ¹⁹

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K max K max e. convert to Joules!

m electron 9.11 10. ³¹

v max 2 K max m electron .

small enough to be nonrelativistic v max 4.61 10= . ⁵

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We are given light of frequencies f.1 and f.2, and the maximum kinetic energies K.1 and K.2 which they produce. We want to find the experimental value for h and φ.

h f 1. K 1 φ Remember, the little square means these are symbolic equations only.

h f 2. K 2 φ

It's very easy to solve these simultaneously for h.

h f 1. K 1 φ h f 2. K 2 φ --- h f 1 f 2. K 1 K 2

In a couple of lines, I'm going to copy this symbolic equation (F2), paste it below (F4), and turn it "on" with "eq".

h K 1 K 2 f 1 f 2

Now plug in numbers K 1 1.97 K 2 0.52 f 1 12 10. ¹⁴ f 2 8.5 10. ¹⁴

This value is in units of eV*s.

You can easily convert it to J*s.

h K 1 K 2 f 1 f 2

h=4.143 10. ¹⁵

Now use the value for h to solve either equation at the start for φ. Use Cut and Paste (F2 and F4) again.

h f 1. K 1 φ Now solve for φ.

φ h f 1. K 1

φ=3.001Since h is in eV*s, units of φ are eV

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To solve this, simply take the equation K max h f. φ

and solve it for h.

K max 1.7

φ 5.4

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λ 175 10. ⁹ c 3 10. ⁸ f c

λ

h K max φ f

h=4.142 10. ¹⁵eV*s

The actual value of h in these units is 4.136x10^-15.

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δf f

G M. c².R

In mks units, G 6.67 10. ¹¹

I'm going to work this problem the straightforward brute strength way. Nothing tricky, but it involves conversions back and forth between f and λ. You might save some time by doing the algebra first on paper.

M 2 10. ³⁰ R 7 10. ⁸ c 3 10. ⁸

λ 500 10. ⁹ the wavelength of the light

f c

λ gives the frequency of the light δf G M. .f

c².R δf=1.27 10. ⁹ The new frequency is less than the original frequency, so

f prime f δf Now calculate the new wavelength:

λprime c

f prime λprime=5.0000106 10. ⁷ The red shift is the amount by which the wavelength changed:

δλ λprime λ

δλ =1.059 10. ¹² meters Or δλ=0.00106 nm.

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Physics 107 Problem 2.52 O. A. Pringle This is just problem 2-51 with different numbers. I will use my 2-51 solution directly here.

In mks units,

G 6.67 10. ¹¹ M 2 10. ³⁰ R 6.4 10. ⁶ c 3 10. ⁸

λ 500 10. ⁹ the wavelength of the light f c

λ gives the frequency of the light

δf G M. .f

c².R δf=1.39 10. ¹¹ The new frequency is less than the original frequency, so

f prime f δf Now calculate the new wavelength:

λprime c

f prime λprime=5.0011583 10. ⁷ The red shift is the amount by which the wavelength changed:

δλ λprime λ δλ =1.158 10. ¹⁰ meters Or δλ=0.116 nm.

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Find the Schwarzschild radius of the sun.

G 6.67 10. ¹¹ M 2 10. ³⁰ c 3 10. ⁸ R S 2 G. .M

c²

R S 2.964 10= . ³ meters

If the sun had less than this radius, it would be a black hole.

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The gravitational potential energy U relative to infinity of a body of mass m at a distance R from the center of a body of mass M is U=-GmM/R.

(a) If R is the radius of the body of mass M, find the escape speed v.e of the body (presumably Beiser means the body of mass m), which is the minimum speed needed to leave it permanently.

To escape, the "body of mass m" must have enough kinetic energy to overcome the gravitational attraction of the body of mass M.

Let me call m the small body and M the large body. The small body starts at the surface of the large body, where the gravitational potential energy is

U (G m. .M) R

and m starts with a kinetic energy of mv^2/2, so our initial total energy is G m. .M

( )

R

1 2.m.v²

After m has just escaped from M, m has used up all of its kinetic energy in escaping, so its kinetic energy is zero. If m had exactly enough energy to just escape, it won't have escaped until it reached a distance of r=∞. At r=∞ the gravitational potential energy is zero (1/r=0 there), so the total energy is zero. Therefore

G m. .M

( )

R

1

2.m.v²=0

"Divide both sides by m, and solve for v to get v 2 G. .M

R

(b) Obtain a formula for the Schwarzschild radius of the body by setting v.e=c, the speed of light, and solving for r.

c 2 G. .M R

Square both sides, solve for R to get R 2 G. .M c²