1.5. FollowingExample 1.1and substituting 200 eV gives:

1 The Wave-Particle Duality - Solutions 1.The energy of photons in terms of the wavelength of light is given by Eq.

(1.5). FollowingExample 1.1and substitutingλ= 200 eV gives:

Ephoton= hc λ = 1240eV∙nm 200nm = 6.2eV

2.The energy of the beam each second is:

Etotal= power time = 100W 1s

= 100J

The number of photons comes from the total energy divided by the energy of each photon (see Problem 1). The photon’s energy must be converted to Joules using the constant 1.602×10 −19 J/eV, seeExample 1.5. The result

is:

Nphotons= Etotal Ephoton = 100J

9.93×10

−19

= 1.01×10

20 for the number of photons striking the surface each second.

3.We are given the power of the laser in milliwatts, where 1mW= 10 −3 W.

The power may be expressed as: 1W= 1J/s. FollowingExample 1.1, the

energy of a single photon is:

Ephoton= hc λ = 1240eV∙nm 632.8nm = 1.960eV

We now convert to SI units (seeExample 1.5):

1.960eV×1.602×10 −19 J/eV= 3.14×10 −19 J

Following the same procedure as Problem 2:

Rate of emission= 1×10 −3 J/s

3.14×10

−19 J/photon

= 3.19×10

15 photons s (Modern Physics with Modern Computational Methods, 3e John Morrison) (Solution Manual all Chapters) 1 / 4

2 4.The maximum kinetic energy of photoelectrons is found usingEq. (1.6) and the work functions, W, of the metals are given inTable 1.1. Following

Problem 1,Ephoton=hc/λ= 6.20eV. For part (a), Na hasW= 2.28eV:

(KE)max= 6.20eV−2.28eV= 3.92eV Similarly, for Al metal in part (b),W= 4.08eVgiving (KE)max= 2.12eV and for Ag metal in part (c),W= 4.73eV, giving (KE)max= 1.47eV.

5.This problem again concerns the photoelectric effect. As in Problem 4, we

use Eq. (1.6):

(KE)max= hc λ −W whereWis the work function of the material and the termhc/λdescribes

the energy of the incoming photons. Solving for the latter:

hc λ = (KE)max+W= 2.3eV+ 0.9eV= 3.2eV

Solving Eq. (1.5) for the wavelength:

λ= 1240eV∙nm 3.2eV = 387.5nm 6.A potential energy of 0.72 eV is needed to stop the flow of electrons. Hence, (KE)maxof the photoelectrons can be no more than 0.72 eV. Solving Eq.(1.6)

for the work function:

W= hc λ −(KE)max= 1240eV∙nm 460nm −0.72eV= 1.98eV

7.Reversing the procedure from Problem 6, we start with Eq. (1.6):

(KE)max= hc λ −W= 1240eV∙nm 240nm −1.98eV= 3.19eV Hence, a stopping potential of 3.19 eV prohibits the electrons from reaching the anode.

8.Just at threshold, the kinetic energy of the electron is zero. Setting (KE)max= 0 in Eq. (1.6), W= hc λ0 = 1240eV∙nm 360nm = 3.44eV 9.A frequency of 1200 THz is equal to 1200×10 12 Hz. Using Eq. (1.10), Ephoton=hf= 4.136×10 −15 eV∙s×1.2×10 15 Hz= 4.96eV 2 / 4

3 Next, using the work function for sodium (Na) metal and Eq. (1.6), (KE)max=Ephoton−W= 4.96ev−2.28eV= 2.68eV

10.We start from Eq. (1.8) for the case ofm= 2:

1 λ =R

1 2 2 − 1 n 2

Now invert the equation and plug in for the Rydberg constant,R:

λ= 1

1.0971×10

5 cm −1

1 4 − 1 n 2

−1

Subtract the fractions by getting a common denominator:

λ= 1cm

1.0971×10

5

n 2 −4 4n 2 −1 Invert the term in the parenthesis and factor out the common factor of 4 λ= 4cm

1.0971×10

5

n 2 n 2 −4

Doing the division, we get Eq. (1.7) for the Balmer formula:

λ= (3645.6×10

−8 cm)

n 2 n 2 −4

11.FollowingExample 1.2,

∆E=−

13.6eV 5 2 −

− 13.6eV 2 2

= 2.86eV

Using Eq. (1.12):

λ= hc ∆E = 1240eV∙nm 2.86eV = 434nm

12.Since the initial state hasm= 2, we can use Eq. (1.7) withn= 4:

λ= (364.56nm)

4 2 4 2 −4

= 486.1nm

To get the energy of the photon, use Eq. (1.5):

Ephoton= hc λ = 1240eV∙nm 486.1nm = 2.551eV 3 / 4

4 13.As in Problem 12, the initial state hasm= 2, so we use Eq. (1.7) with

n= 3:

λ= (364.56nm)

3 2 3 2 −4

= 656.2nm 14.From Figure 1.6, the ionization energy of a hydrogen atom in then= 2 state is−3.4eV. So it takes a photon of 3.4eVto just ionize this atom. To

get the wavelength of light, just invert Eq. (1.5):

λ= hc Ephoton = 1240eV∙nm 3.40eV = 364.7nm

15.Starting with Eq. (1.5) with a wavelength of 200 nm:

Ephoton= hc λ = 1240eV∙nm 200nm = 6.20eV From Figure 1.6, a hydrogen atom in then= 2 state has the electron bound with potential energy (P E) =−3.4eV. FollowingExample 1.3, (KE) = 6.20eV−3.40eV= 2.80eV

16.Starting with Eq. (1.5) with a wavelength of 45 nm:

Ephoton= hc λ = 1240eV∙nm 45nm = 27.6eV For a hydrogen atom in the ground state, the electron is boundwith potential energy (P E) =−13.6eV. FollowingExample 1.3, (KE) = 27.6eV−13.6eV= 14.0eV

To find the electron’s velocity, convert to SI units (seeExample 1.5):

(KE) = 14.0eV∙

1.6×10

−19 J 1eV

= 2.24×10

−18 J FromAppendix A, an electron has massm= 9.11×10 −31 kg. Using the well known formulaKE= (1/2)mv 2

, and solving forv:

v= r 2(KE) m = s

4.48×10

−18 J

9.11×10

−31 kg

= 2.22×10

6 m s 17.From Figure 1.6, we see that the first transition for the Lymanseries is betweenn= 2 ton= 1, and similarly we can get the transitions for the

Balmer and Paschen series. Using Eq. (1.8):

1 λ =R

1 m 2 − 1 n 2

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