E41-1.2 Using the classical relation between the momentum and the kinetic energy,show that the de Broglie wavelength of an electron can be written (a) asλ = 1.226√

Knm, in which K is the kinetic energy in electron volts, or (b)

λ = 1.5 VV nm. Where V is the accelerating potential in volts. [Constants

would be given]

Solution: As K.E. has been given in eV, we may follow this method:λ = h

p = hc√2mc2K

E = mc2 = (9.11×10−31kg)×(9.0×1016(m/s)2) = 81.99×10−15kg.m2.s−2(J) =81.99× 10−15(J)× (6.242× 1018eV) = 511keV.

h = (6.63× 10−34J.s)× (6.242× 1018[eV]) = 41.384× 10−16[eV.s]

hc = (41.384 × 10−16eV.s) × (3.0 × 108m/s) = 123.9 × 10−8eV.m =1240eV.nm

(a) λ = h/p and p2/2m = K, then λ = hc√2mc2K

λ = 1240eV nm√2(511keV)K

= 1.226nm√K

(b)(1/2)mv2 = eV , so v =√

2eVm ⇒ λ = h√

2emV= 6.63×10−34J.s√

2(1.6×10−19)×(9.11×10−31kg)V

λ =√

43.9569×10−68

2×1.6×9.11×10−50V nm =√

1.5VV nm

E41-1.4 The wavelength of the yellow spectral emission line of sodium is 589 nm.At what kinetic energy would an electron have the same de Broglie wave-length?

Soln: p = h/λ and p2/2m = K, soK = (hc)2/(2mc2λ2). K = (1240eV nm)2/2(511×103eV)(589nm)2 = 4.34× 10−6eV

E41-1.5 If the de Broglie wavelength of a proton is 0.113 pm, (a) what is the speedof the Proton and (b) through what electric potential difference would theproton have to be accelerated from rest to acquire this speed? [Mass ofproton = 1.67×10−27 kg

(a) p = h/λ. First Calculate for proton, mc2 = (1.67 × 10−27kg)(9.0 ×109m.s−2)(6.242× 1018eV ) = 93.8× 107eV = 938× 106eVSpeed, v = h/(mλ) = c(hc)/(mc2λ)= c(1240× 10−9eV.m)/[(938× 106eV )(0.113× 10−12] = 0.0117c

Then v = 3510 km/s.

(b) Kinetic energy, K = (1/2)mv2 = (1/2)(mc2)v2/c2 = (1/2)(938 ×106eV)(0.0117c)2/c2 = 0.0642× 106eV Potential =64.2 kV.

E41-1.5 Singly charged sodium ions are accelerated through a potential differenceof 325 V. (a) What is the momentum acquired by the ions? (b) Calculatetheir de Broglie wavelength.

(a) For a singly charged Na atom, no. of proton = no. of neutron 11:mass (m)22 (of proton mass). Also being a singly charged, you think of

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only one proton extra.p =

√2mc2K/c2.

p =√

2(22)(938× 106eV/c2)(325eV ) = 3662× 103eV/c

(b) λ = h/p, so λ = hc/pc = (1240eV nm)/[(3662× 103eV/c)c] = 338.8×10−6 nm

Sample Problem: 46-6 An electron total energy E = 5.0 eV approaching a potential barrier (fig.1),whose height Uo = 6.0 eV. The barrier thickness L = 0.7 nm. (a) What isthe de Broglie wavelength of the incident electron? (b) What transmissioncoefficient T follows fromT = 16 E

Uo

(1− E

Uo

)exp(−2k′L)

(c) What would be the transmission coefficient if the barrier thicknesswere reduced to 0.35 nm? If its height were increased to 7.0 ev? If theincident particle were a proton?

Soln.: (a) Energy in eV, you should covert to Joule.Total energy E = K,λ = h/p = h√

2mK

= 6.63×10−34J.s2(9.11×10−31kg)(5eV )(1.6×10−19J/eV )

= 6.63×10−34√145.76×10−50

= 0.549× 10−9m = 0.55nm

(b) To find T , among other things we need k′ =2π√

2m(Uo−E)

h

=(2π)√

2(9.11×10−31kg)(6.0−5.0eV )(1.6×10−19J/eV )

6.63×10−34J.s = 5.12× 109m−1

k′L = (5.12× 109m−1)(0.7× 10−9m) = 3.58

T = 16 EUo

(1− E

Uo

)exp(−2k′L)

= T = 16( 56 )(1− 5

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)exp(−2(3.58)) = 2.222(exp(−2(3.58))) = 1.73×10−3

This means out of 100,000 electrons, only 173 can tunnel through thebarrier.

(c) Rest of calculation is easy.

46-4.22 An atom in an excited state has a lifetime of 12 ns; in a second excitedstate the lifetime is 23 ns. What is the uncertainty in energy for thephoton emitted when an Electron makes a transition between these twostates?

Soln.: Find uncertainty for each states and add to find total.Heisenberg principleFor h convert to eV.s∆E1 = h/(2π.∆t) = (4.14× 10−15eV.s)/2(12× 10−9s) = 5.49× 10−8 eV:and∆E2 = (4.14 × 10−15eV.s)/2(23 × 10−9s) = 2.86 × 10−8 eV The sum isET = 8.35× 10−8 eV.

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