Tutorial 8: Solutions

Transcription

1 Tutorial 8: Solutions 1. * (a) Light from the Sun arrives at the Earth, an average of m away, at the rate Watts/m of area perpendiular to the diretion of the light. Assume that sunlight is monohromati with a frequeny Hz. (i) How many photons fall per seond on eah square metre of the Earth s surfae diretly faing the Sun? (ii) What is power output of the Sun, and how many photons per seond does it emit? (b) A 1000 Watt radio transmitter operates at a frequeny of 880 khz. How many photons per seond does it emit? (a) (i) We have E hν so the energy in eah photon is E J. The number of photons in the total energy of J is therefore given as n (a) (ii) The power output of the Sun is: (b) The number of photons is: (1) P 4π ( ) W W () n (3). * (a) Illuminating the surfae of a metal alternately with light of wavelengths λ µm and λ 0.54µm, it was found that the orresponding maximum veloities of the photoeletron differ by a fator of. Find the work funtion of the metal. (b) The maximum energy of photoeletrons from Alumunium is.3 ev for radiation of 000Å and 0.90 ev for radiation of 580Å. Use these data to alulate Plank s onstant and the work funtion of Alumunium. (a) We have 1 mv 1 hν 1 W (4) 1 mv 1 4 hν W (5) 1

2 Therefore W 4 ( 3 h ν ν ) 4 ( h 1 ) (6) λ λ 1 Subsituting the values, we get W J (7) (b) We have T max hν W. Therefore h 10 W (8) h 10 W (9) On solving these two equations for h and W one gets h ; W 4.83 ev (10) 3. * (a) A Compton effet experiment is performed in suh a way that the sattered photon and the reoil eletron are deteted only when their paths are at right angles (θ + φ π ). (i) Show that under these onditions λ λ os φ (ii) Show that the energy of the sattered photon is m 0. (iii) Find the energy of the reoil eletron. (iv) What is the minimum energy of the inident photons for whih the experiment an be done? (b) Show that when a free eletron is sattered in a diretion making an angle θ with the inident photon in a Compton sattering, the kineti energy of the eletron is: E k hν (α os θ) [(1 + α) α os θ] (11) where α hν m 0. (a) We assume the angle of sattering of the photon as φ and that of the reoil eletron as θ. Therefore, onservation of energy gives hν + m 0 hν + p + m 0 4 (1)

3 where p is the momentum of the reoil eletron and m 0 is the rest mass of the free eletron. Conservation of momentum along the horizontal and vertial diretions give (i) Sine θ + φ π hν hν os φ + p os θ (13) hν sin φ p sin θ (14) we have, from the onservation of momentum equations hν hν os φ + p sin φ (15) hν sin φ p os φ (16) Multiplying the first of the above two equations by osφ and the seond by sinφ and subtrating them, one gets whih, in terms of wavelength is ν ν os φ (17) λ λ os φ (ii) From the momentum onservation equations obtained after using θ + φ π easily obtain (18) we an p hν sin φ (19) Using this in the energy onservation equation gives hν + m 0 hν + (hν) sin φ + m 0 4 (0) Writing sinφ 1 ( ) ν ν we get hν + m 0 hν + whih gives, on a rearrangement and squaring (hν) (hν ) + m 0 4 (1) ( hν hν + m 0 ) (hν) (hν ) + m 0 4 () This equation an be rewritten as h (ν ν ) ( hν m 0 ) 0 (3) Sine ν ν, we have the energy of the sattered photon as hν m 0 (4) 3

4 (iii) The energy of the reoil eletron is obtainable from hν + m 0 hν + E reoil (5) Sine hν m 0, E reoil hν. (iv) The minimum energy of the inident photons required for the experiment to take plae is given as hν m 0, i.e. when p 0. In this ase however, there is no hange of the frequeny of the inident photon and the eletron also remains at rest. (b) The kineti energy of the sattered eletron is E k p + m 0 4 m 0 (6) Using the fat that p + m 0 4 hν hν + m 0 (from energy onservation) we an rewrite E k as E k h (ν ν ) h m 0 λ sin φ λ + h m 0 sin φ where we have onverted to wavelengths and use λ λ h m 0 (1 os φ). Now we will need to onvert this in terms of θ. For this we need to know the relation between θ and φ. Using the onservation equations for momenta given as we get From the Compton sattering formula we have Therefore, the equation for otθ beomes (7) hν hν os φ + p os θ (8) hν sin φ p sin θ (9) ot θ ν 1 otφ (30) ν sin φ ν 1 + hν (1 os φ) 1 + α (1 os φ) (31) ν m 0 ot θ (1 + α (1 os φ)) Mutiplying by sinφ on both sides and after a bit of algebra one obtains 1 otφ (3) sin φ otθ (1 + α) tan φ (33) 4

5 The above expression for otθ is equivalent to sin φ ot θ (1 + α) + ot θ Substituting this bak in the expression for E k and after some straightforward algebra one obtains E k αhν os θ (1 + α) α os θ (34) (35) 4. (a) Find the de Broglie wavelength of a 1 mg grain of sand blown by the wind at a speed 0 m/s. * (b) Calulate the de Broglie wavelength of an eletron when its energy is 1 ev, 100 ev, 1000 ev. At what value of the kineti energy is the de Broglie wavelength of an eletron equal to its Compton wavelength? () Show that the ratio of the de Broglie wavelength to the Compton wavelength of the same partile is equal to ( v) 1. (a) The de Broglie wavelength here is: λ h mv m (36) (b) The de Broglie wavelength an be found from the formula λ h mv h me. For 1 ev, 100 ev and 1000 ev, the de Broglie wavelengths are 1.8 Å, 1.8 Åand 0.39 Å. Using h h one gets v m 0. This results in v 1 m 0 mv m (using m q m 0 ). Therefore, 1 v the energy of the eletron is E m 0 m MeV. Therefore the KE E m 0 0.1MeV. () This follows from the previous problem. We have λ db m 0 λ m v v 1 (37) 5. * (a) A parallel beam of eletrons aelerated by a potential differene of 5 Volts falls normally on a diaphragm with two narrow slits separated by a distane, 50 µm. Calulate the distane between neighboring maxima of the diffration pattern on a sreen loated at a distane, 100 m from the slits. 5

6 (b) A parallel beam of monoenergeti eletrons falls normally on a diaphragm with narrow square slit of width 1 mirometre. Find the veloity of the eletrons if the width of the entral diffration maximum formed on a sreen loated 50 m from the slit is equal to 0.36 mm. (a) The de Broglie wavelength of eletron is λ h p h me E h me ev h me ev m e 0.501MeV is the rest mass energy of eletron. h Js, V 5V. So λ.47angstrom. Separation between two neighboring maximum of diffration pattern on the sreen is where, L 100m and d 10µm. L θ L λ d 4.9µm (b). Width of the entral diffration maximum is λl a 0.36mm where, λ is de Broglie wavelength of the eletron, L 50m and a 1µm. So λ 3.6 Angstrom. Hene speed of the eletron is where, m e kg. v h m e λ 106 m/s 6. * (a) The position and momentum of a 1 kev eletron are simultaneously determined. If the position is loated to within one Angstrom what is the perentage of unertainty in its momentum? 6

7 (b) Verify that the unertainty priniple an be written in the form L θ where L is the unertainty in the angular momentum of a partile and θ is the unertainty in its angular position. * () An atom in an exited state has a lifetime of 1 nanoseonds and in a seond exited state the lifetime is 3 nanoseonds. What is the unertainty in energy of the photon when an eletron makes a transition between these two states? (a). Minimum unertainty in momentum is Momentum of 1 KeV eletron is h p 4π x kg m/s p m e E kg m/s So p p 3.08% (b). The unertainty relation between angular momentum and angular position an be verified from the unertainty relation between linear momentum and position. x p h 4π Consider a partile with linear momentum p moving on a irle of radius r. The partiles angular momentum is given by, L pr.in moving a distane x on the irle the partile sweeps out an angle θ x/r. Then for a fixed radius r, θ x/r and L r p. So, L θ x p h 4π (). We know the the energy-time unertainty relation, E t h/4π. For t τ 1 1ns, E J and for t τ 3ns, E J. So the minimum unertainty in energy of the photon is ( E 1 + E ) J. 7. (a) A partile is loated in a one dimensional square potential with infinitely high walls. The width of the well is a. Find the normalised wave funtions of the stationary states of the partile taking the midpoint of the well as the origin of the x oordinate. 7

8 (b) A partile is onfined in a one dimensional box loated between x 0 and x L. The potential is infinity at the walls and zero inside. Determine the probability of finding the partile in a region of width x 0.1L at eah of the following loations when it is in its ground state: x 0, 0.5L, 0.5L.0.75L,L. Repeat this exerise for the first exited state. (a). For a partile loated in a one dimensional potential, we solve the time independent Shrödinger equation, ħ d u(x) + V (x)u(x) Eu(x) m dx where, ψ(x,t) e iet/ħ u(x) is the wave funtion of the stationary states of the partile. Now, V (x) for x a/ and x a/; V (x) 0 for a/ < x < a/. So, for x a/ and x a/ u(x) 0. For a/ < x < a/, the above equation beomes d u(x) dx + k u(x) 0 where, k me/ħ. The general solution of this equation is u A os kx + B sin kx Using the boundary onditions, u( a/) u(a/) 0, we have two equations A os(ka/) + B sin(ka/) 0 A os(ka/) B sin(ka/) 0 For non trivial solution of A and B, os(ka/) sin(ka/) 0 (determinent of the oeffiient matrix is zero). So sin(ka) 0, or, ka nπ, where n is an integer.for even n, A 0 and for odd n, B 0. So the solution for u is ( nπx ) u B sin ( a nπx ) A os a Now using the normalisation ondition, we have a/ a/ n, 4, 6,... n 1, 3, 5,... ( nπx ) B sin dx 1 a performing the integration we get B /a. Similarly, A /a. So finally the normalised wave funtion is 8

9 where, E n n π ħ ma. ( nπx ) ψ(x,t) a e ient/ħ sin a ( nπx ) a e ient/ħ os a n, 4, 6,... n 1, 3, 5,... (b). The probability of finding the partile in a region of width x, at the loation x is P x ψ(x,t) (38) where ψ(x,t) normalised wave funtion. In this problem, ψ n (x,t) L e ient/ħ sin ( ) nπx L. Then the expression for P beomes For ground state n 1. P πx x sin L L (39) Thus the probabilities at x 0, 0.5L, 0.5L, 0.75L, L are 0, 0.1, 0., 0.1, 0 respetively. One an repeat this easily for the n eigenfuntion. 8. * (a) An eletron is trapped in a region of width L 10 Angstroms. What are the energies of the ground state and the first two exited state? An eletron in the first exited state de-exites to the ground state. What is the wavelength of the emitted radiation? (b) The wavefuntion of a partile onfined in a one dimensional box, between x 0 and x L, is given to be ψ(x,t) 3 5 ψ 1(x,t) ψ (x,t) (40) where ψ 1 (x,t) and ψ (x,t) are the ground and first exited state wavefuntions of the partile in a box. What are the possible outomes and their probabilities if the energy of the partile is measured? (a). The energies of an eletron trapped in a region of width L 10 Angstrom are E n n π ħ ml for ground state n 1 and for first exited state n. So E eV and E 1.5eV. The wave length of the emitted radiation for transition first exited state to ground state is λ h E E 1 1.1µm 9

10 (b). If the energy of the partile is measured then after every measurement the outome will be either ground state energy or the first exited state energy, sine ψ(x,t) is a ombination of ground state and first exited state. E 1 π ħ ml and E π ħ ml. If the measurement is performed several times then the probability of geting the ground state energy is P /5 and probability of finding the first exited energy is P /5, where 1 3/5 and 4/5. 10