7/27/2019 Midterm 1 SOLNS

1/4

Solutions for Midterm #1

PHYS 17 - Winter 2014

Problem 1

Estimate the surface temperature of the Sun from the following information. The Suns radius isgiven by RS= 7.0 10

8 m. The average Earth-Sun distance is R= 1.5 1011 m. The power perunit area (at all frequencies) from the Sun is measured at the Earth to be 1400 W/m2. Assumethat the Sun is a blackbody.

Solution:

Total power per unit area is given by

etotal(Rs) =T4

(the notation here is that etotal is a function of the radius of the sun). Total power emitted fromthe Sun is given by

P =etotal(Rs)(4R2s) =T

4s(4R

2s)

All of this power is radiated uniformly through the area of the Sun. This relationship still holds ifwe go out a distance R, although our sphere will be larger. Therefore

etotal(Rs) =etotal(R)R2

R2s

But this is still equal toT4. Solving for T:

T =etotal(Rs) R2

R2s1/4

= 5800K

1

7/27/2019 Midterm 1 SOLNS

2/4

Problem 2

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered raysare detected at 30 relative to the incident rays, find(a) the Compton shift, in wavelength, at this angle(b) the energy of the scattered x-ray(c) the energy of the recoiling electron

Solution:

(a) Use the Compton scattering formula:

= 0

= h

mec(1 cos )

= 6.623 1034Js

9.11 1031

kg 3 108

m/s

(1 cos30)

= 3.25 1013m

(b) You know the energy of the original photon, solve for its wavelength, add the Compton shift,and convert back into energy.

E0 = hc

0= (300keV)(1.602 1019J/eV)

= (6.63 1034Js)(3 108m/s)

0

0 = 4.14

1012

m = 0+

= 4.465 1012m

E = hc

= (6.63 1034Js)(3 108m/s)

4.465 1012m= 278 keV

(c) Find the change in the photons energy and by conservation of energy that will be the electronsrecoil energy.

Ke = E0

E= 300

278 = 22 keV

2

7/27/2019 Midterm 1 SOLNS

3/4

Problem 3

(a) Find the frequency in Hz of a photon emitting when a hydrogen atom makes a transition froma state with n = 30 to one with n= 29.(b) Find the orbital frequency (i.e. the frequency of revolution) of the electron in Hz in its initialstate.(c) Why the value of the emission frequency is so close to that of the orbital frequency?

Solution:

(a) Use the Hydrogen transition formula and solve for frequency

hf = E0

1n2i

1

n2f

f = 13.6 eV

4.15 105 eV s

1302

1

292

= 2.6 1011

Hz

(b) Compare orbital force to Coulomb Force

ke2

r2 =

mv2

r

v =

ke2

mr

forbital = v

2r

= 1

2rke2

mr

= 1

2(0.5A)

(9 109m/F)(1.6 1015C)2

(9.1 1031)(0.5A)

= 2.9 1011 Hz

Alternate forms offorbital wherer was replaced withn22/mke2 were also acceptable.

(c) They are so close to each other because n = 30 is a large quantum number. Asn , thecorrespondence principle says that quantum mechanical results must approach classical results.

3

7/27/2019 Midterm 1 SOLNS

4/4

Problem 4

SEE PLOT.

The following figure shows the stopping potential versus incident photon frequency for the photo-electric effect for molybdenum. Use these data point to find(a) the ratio h/e(b) the work function.

Solution:

(a)eVs= hf

Vs =hf

e

e

The slope is then h/e. Pick two spots and compute:

he

= 16.2 65 1015 2 1015

= 4.2 1015 V s

(b) Take one point, such as (3 1015, 8.2). Solving this:

8.2 = (4.2 1015)(3 1015)

= 4.4eV

4