Solución 1

a) m=46g beis

λ p=hp= hmv

= 6.63 x10−34 J∗s4.6 x10−2kg∗30m / s

=4.8014 x10−34m

b) eleectron

λ p=hp= hmv

= 6.63x 10−34 J∗s9.1x 10−31∗107m /s

=7.2857 x10−11m

c) particula de polvo

λ p=hp= hmv

= 6.63 x 10−34 J∗s0.001kg∗20m/ s

=3.3115 x 10−32m

Solución 2

E=400keV E= h2

8n L2n2

L2= h2

8nEn2

he=1240eV∗nm ne2=0.513MeV m=1.67*10 a la-27kg

L=√ (hc )2

8mc2 En2=√ (1240eV∗nm )2

8 (5.11 x5eV )400keV22=1.039nm=1.039 x10−12 mal resultado

Solución 3

Δt=10−8 s

h'= h2π

=6.581x 10−16 eV

ΔE Δt ≥h'

2Principio de incertidumbre

ΔE ≥h'

21Δt

ΔE ≥6.581 x10−16eV s

2∗10−8 s

ΔE ≥5.27∗10−27J=3.29 x 10−8 eV⏟

Solución 4

Longitud de onda corta

λmin nF=1 , ni=4

1λ=Rm(

1

nF2− 1

ni2)

1/ λmin=Rm(1/1−1/ inf )

λmin=1 /(10.973 x106m−1)

λmin=911˙A=9.11 x10−8m

Longitud de onda larga

λmax nF=1, ni=2

1λ=Rm(

1

nF2− 1

ni2)

λmax=1

[(10.973 x 106m−1) (11−14 )]=1.21∗10−7m=1215 A

Solución 5

n f=2 , ni=5

n f=2=−3.4eV n i=5=−0.54eV

λ=hcE

=(4.135x 10−15 eV s)(3 x 108m / s)

−3.4 eV−0.54eV=4.34 x 10−7=4340 A

1λ=Rm(

14− 125

)

λ= 1

[Rm( 14−125 )]

=4.339 A

Solución 6

λ=1240eV∗nmΔE

ΔE=E z−E f=z2E0(1

nF2− 1ni2)

Para z=1y E0=13.6eV

ΔE=13.6eV ( 1nF2− 1

ni2)

λ=1240eV∗nmΔE

= 1240eV∗nm

13.6eV (1

nF2 −

1

ni2 )

= 91.17nm

(11−1

102)=92.121nm=921 A=9.2 x10−8m

Solución 7

a)

r=n2a0

n2=0.01mma0=0.0529nm

n=√ ra0

=√ 10000nm0.0529nm=435

b)

E=13.6 eVn2

=13.6eV4352

=7.187 x10−5 eV