PHYSICS 61A WINTER 2012Midterm Exam: Feb 7, 2012, 11-12:20pm

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Total number of pages: 7

NAME:

1. (20pt)

2. (15pt)

2. (15pt)

3. (15pt)

4. (15pt)

5. (20pt)

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

TOTAL: (100pt)

1

1. Electrons of energy 12.2 eV are fired at the hydrogen atoms in a gas discharge tube. Deter-mine the wavelengths of the lines that can be emitted by the hydrogen. [20 points]

Solution: Maximum energy that can be absorbed by a hydrogen atom is equal to electronenergy, 12.2 eV. Absorption of this energy could excite atom into an energy state En, assumingatom initially at the ground state.

En = E1 + 12.2 eV = −13.6 ev + 12.2 eV , (1)

En = −E1

n2. (2)

Therefore,

−14.4 eV = −13.6 eV

n2(3)

orn = 3.12 . (4)

Since n must be an integer, n = 3. There are three possible wavelengths, corresponding tothree different transitions.

3→ 2 :1

λ= R

(1

22− 1

32

)= (0.011 nm)−1(1/4− 1/9) , (5)

2→ 1 :1

λ= R

(1

12− 1

22

)= (0.011 nm)−1(1/1− 1/4) , (6)

3→ 1 :1

λ= R

(1

12− 1

32

)= (0.011 nm)−1(1/1− 1/9) . (7)

2

2. If the maximum energy imparted to an electron in Compton scattering is 60 keV, what is thewavelength of the incident photon? [15 pt]

Solution:

Suppose the incident photon has wavelength λ0 and energy E0, while the scattered photonhas wavelength λ and energy E.

The maximum energy goes to the electron when θ = 180o. In this case,

λ− λ0 = ∆λ =2h

mc= 0.00486 nm . (8)

The electron gains maximum kinetic energy,

∆E = E0 − E =hc

λ0− hc

λ0 + ∆λ. (9)

Solve for unknown λ0 in the quadratic equation in

λ20 + ∆λ · λ0 −hc∆λ

∆E= 0 . (10)

There are two solutions to the equation:

λ0 = −∆λ

2±

√(∆λ

2

)2

+hc∆λ

∆E. (11)

Only the plus sign gives the physical solution. Thus,

λ0 = 0.00788 nm . (12)

3

3. Consider a proton confined to a region of typical nuclear dimensions, about 5 fm.

(a) Use the uncertainty principle to estimate its minimum possible kinetic energyin MeV,assuming that it moves only in one dimension. [9 pt]

(b) How would your result be modified if the proton were confined in a three-dimensionalcube of side 5 fm? [6 pt]

Solution:

(a)

Kmin =(h̄c)2

2ma2=

(h̄c)2

(2mc2a2)(13)

=(197 MeV · fm)2

2× (938 MeV)× (5 fm)2(14)

= 0.85 MeV . (15)

(b)

K =p2x + p2y + p2z

2m(16)

ThusKmin = 3× 0.85 MeV = 2.5 MeV . (17)

4

4. A surface has light of wavelength λ1 = 550 nm incident on it, causing the ejection of photo-electrons for which the stopping potential is Vs1 = 0.2 eV. Suppose the radiation of wavelengthλ2 = 200 nm were incident on the surface. Calculate

(a) the stopping potential Vs2, [7 points]

(b) the work function of the surface, and [4 points]

(c) the threshold frequency for the surface. [4 points]

Solution:

(a)

eV = KEmax = hf −W (18)

⇒ e(Vs2 − Vs1) = h(f2 − f1) (19)

⇒ Vs2 = Vs1 +h

e(f2 − f1) = Vs1 +

hc

e

(1

λ2− 1

λ1

)(20)

= 0.20 + 1240(1/200− 1/550) = 4.15 V (21)

(b)

W =hc

λ1− eVs1 =

1240

550− 0.2 = 2.05 eV . (22)

(c)hf = W (23)

⇒ f =W

h=

2.07× (1.602× 10−19)

6.63× 10−34= 500 THz . (24)

5

5. A photon (λ = 0.200 nm) strikes an electron at rest and rebounds at an angle of 120o to itsoriginal direction.

(a) Find the speed of the photon after the collision. [5 points]

(b) Find the wavelength of the photon after the collision. [10 points]

Solution:

(a) speed of light in vacuum = c

(b)

λ′ = λ+h

mc(1− cosφ) (25)

= 2× 10−10 m +6.63× 10−34 J · s

(9.1× 10−31 Kg)(3× 108 m/s)× (1− cos 120o) (26)

= 0.2036 nm . (27)

6

6. An unusually long-lived unstable atomic state has a lifetime of 1 ms.

(a) Roughly what is the minimum in its energy? [10 pt]

(b) Assuming that the photon emitted when this state decays is visible (λ ' 550 nm), whatare the uncerntainty and fractional uncerntainty in its wavelength? [10 pt]

Solution:

(a)

∆E ≥ h̄

2∆t=

6.58× 10−16 eV · s2× 10−3 s

= 3.3× 10−13 eV . (28)

(b)

λ =hc

E⇒ ∆λ =

(hc

E2

)∆E =

(λ2

hc

)∆E . (29)

Therefore∆λ = 8× 10−11 nm , (30)

and∆λ

λ= 1.5× 10−13 . (31)

7