Why was Quantum Mechanics Developed?

1 Why was Quantum Mechanics Developed?

1.1 Calculate the energy of a photon of 253.6 nm light. State the energy in eV (1eV = 1.6×10 −19 J).We obtain E=hc/λ =

6.626×10

−34 Js×3×10 8 ms −1

253.6×10

−9 m

= 7.83832×10

−19 J =

7.83832×10

−19

1.6×10

−19 eV = 4.89895 eV.

1.2 Calculate the wavelength of a quantum of electromagnetic radiation with energy 4000 eV.We haveλ=hc/E. We have hc= 6.626×10 −34 Js×3×10 8 ms −1 .SinceEis in eV we expresshcin units of eV. The conversion factor is 1J = 1/(1.6×10 −19

• eV. So,

hc=

6.626×10

−34 eV×3×10 8 m

1.6×10

−19

= 1242.375×10

−9 eVm = 1242.375 eVnm.Then the wavelength of a quantum of electromagnetic radiation with energy 4000eV is λ= 1242.375 eVnm 4000 eV

= 3.106×10

−10 m = 3.106˚A.1 Solutions Manual for Quantum Mechanics I The Fundamentals, 2e by S. Rajasekar, R. Velusamy (All Chapters) 1 / 4

2Why was Quantum Mechanics Developed?

1.3 The Planck’s radiation formula for an ideal black body is E(ν)dν= 8πν 2 dν c 3 hν e hν/kBT −1 , whereE(ν)dνis the energy density in the frequency interval be- tweenνandν+ dν. Determineλmax.Substitution ofν=c/λ, dν=−(c/λ 2 )dλandE(ν)dν=−p(λ)dλ in the given formula. The result is p(λ) = 8πhc λ 5 1 e hc/(λkBT) −1 .The wavelengthλat whichp(λ) becomes maximum can be found from the condition dp(λ)/dλ= 0. We get λ= hc kBT e hc/(λkBT) 5 Γ e hc/(λkBT) −1 ∆ = hc kBT 1 5 θ

• +hc/(λkBT)

hc/(λkBT)

That is, λmax= 0.25 hc kBT .

1.4 Consider the exercise 1.3. Examine the spectral shift accordingto Wien’s displacement lawλmaxT= constant if the value ofhis doubled.We got the result λmax= 0.25 hc kBT .Whenhbecomes 2hthenλmaxwould become doubled. Then the solar surface temperatureTSalso change. We have 4πR 2 σT 4 S =G,

σ= 2π

5 k 4 B /(15c 2 h 3

• andM/R≈2k

2 B q 4 /(Gh 2

• SinceR∝h

2 , σ∝h −3 andG∝h −3 . We haveTS∝h −1 . Therefore, the peak wavelength of the sun will follow thatλmax,sun= 0.25hc/(kBT)∝ h 2 . As a result the main radiation from the sun would be shifted to the infrared region. As pointed out in [P.K. Yang,Eur. J. Phys.

37:055406, 2016] the creatures on earth might develop eyes capable

of detecting infrared radiation. 2 / 4

3 1.5 A light of wavelength 1000 nm produces photoelectrons from a metal. The threshold potentialV0is 0.5 V. Calculate the work func- tionφ0.The work functionφ0is given by φ0= hc λ − |e|V0.hcis calculated as hc= 6.626×10 −34 Js×3×10 8 ms −1 =

6.626×10

−34

×3×10

8

1.6×10

−19 eVm = 1242.375 eVnm.An electron accelerated by 1 V will have 1 eV of kinetic energy.Therefore, an electron accelerated by 0.5 V will have the kinetic energy of 0.5 eV. Then φ0= 1242.375 eVnm 1000 nm −0.5 eV = 0.742375 eV.

1.6 Consider the previous exercise. Calculate the threshold wavelength λc.We obtain λc= hc E = 1242.375 eVnm 0.742375 eV = 1674 nm.

1.7 Calculate the frequency of light required to produce electrons of ki- netic energy 5 eV from illumination of a material whose work func- tion is 0.5 eV.According to conservation of energy Energy of a photon = K.E.of an emitted electron + Work function.We obtain the energy of the incident light as E= 1 2 mv 2 max+φ0= 5eV + 0.5eV = 5.5 eV.Then ν= E h = 5.5 eV

6.626×10

−34 Js .Since 1 eV = 1.6×10 −19 J we get ν=

5.5×1.6×10

−19 J

6.626×10

−34 Js

= 1.3281×10

15 Hz. 3 / 4

4Why was Quantum Mechanics Developed?

1.8 Light of wavelength 500 nm is incident on a metal with work func- tion (φ0) of 0.75 eV. What is the maximum kinetic energy (Kmax) of the ejected photoelectrons?Kmaxishν−φ0. We obtain Kmax=hν−φ0= hc λ

−φ0

=

6.626×10

−34 Js×3×10 8 ms −1

500×10

−9 m

−0.75×1.6×10

−19 J

= 2.7756×10

−19 J.

1.9 Suppose you are standing in front of a 40 W incandescent light bulb

• m away. If the diameter of your pupils is about 2 mm, about how

many photons,n, enter your eye every second?Intensity at the eye is I= 40πd 2 4πr 2 = 10d 2 r 2 W, wheredis the diameter of the pupil andris the distance between the eye and light source. We get I=

10×(0.002)

2 5 2

W = 1.6×10

−6 Js −1 .The number of photons that enters the eye every second is obtained as n= I hν =

1.6×10

−6 Js −1

6.626×10

−34 Js×ν =

0.2414729×10

28 ν s −2 .Ifνis stated in Hz then n=

2.414729×10

27 ν s −1 .

1.10 Calculate the minimum voltage required for an electron to give all of its energy in a collision with a target and produce an electromag- netic radiation of wavelength 0.05 nm.When an electron is accelerated through a potential differenceV, it acquires an energy|e|V. If this energy goes into producing a photon of energyEwith associated wavelengthλ, we getE=hc/λ. So E=

6.626×10

−34 Js×3×10 8 ms −1

0.05×10

−9 m =

6.626×10

−34

×3×10

8

1.6×10

−19

×0.05×10

−9 eV = 24847.5 eV.

• / 4