html5-img
1 / 11

1. Photoelectric effect

ph oton. A. ele c tron. K. A. V. P oten t iome ter. 1. Photoelectric effect. Hertz, The photoelectric effect is the emission of electrons when light strikes a surface .

jorn
Download Presentation

1. Photoelectric effect

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. photon A electron K A V Potentiometer 1. Photoelectric effect Hertz, The photoelectric effect is the emission of electrons when light strikes a surface.

  2. Photocurrent i for a constant light frequency f as a function of the potential VACof the anode with respect to the cathode.

  3. =the minimum energy needed to remove an electron from thesurface • Kmax= ½mv2max = maximum kinetic energy of photoelectrons where h is a universal constant called Planck's constant. Vo = stopping potential

  4. Contohsoal1 Fluksenergimataharisampaikebumiadalah 1,0 x 103 W/m2.. Berapakahcacahfoton yang sampaidibumi per m2tiapdetiknyajikacahayamataharimempunyaipanjanggelombang rata-rata 5000 ? Jawab Energitiapfotondapatdihitung E = h = hc/ = (6,62 x 10-34 Js) (3 x 108 m/s ) /5 x 10-7 m = 4,0 x10-19 J Fluksenergifoton yang sampaidibumi 1,0 x 103 W/m2 =1,0 x 103 Js-1m2 Iniberartienergi yang sampaidibumi per m2tiapdetiknya 1,0 x 103J. Olehkarenaenergitiapfotonadalah 4,0 x10-19 J makacacahfoton yang sampaidibumitiap m2tiapdetikadalahdetiknya 1,0 x 103 J / 4,0 x10-19 J = 2,5 x1021foton.

  5. h Ek B Contohsoal 2 Energi yang diperlukanuntukmelepaselektrondarilogamplatina 9,9 x 10-19 J. Berapakahfrekuensiambangcahayauntukmenghasilkanfotoelektrondariplatinatersebut? Jawab Energifoton yang dijatuhkanpadalogamplatinasebagiandigunakanuntukmelepasikatanelektron (B) dansebagian lain menjadienergikinetikfotoelektron (Ek) GambarFotonmenumbuklogamplatina. Energifoton yang besarnyasamadenganenergi yang dibutuhkanuntukmelepas elektrondariikatannyadisebutenergiambang. Frekuensi yang berkaitandenganenergiambangdisebutfrekuensiambang. Denganmenggunakan E = -B + hatau h = E + B dimana E adalahenergikinetikelektron. Jika E = 0 maka h = B disebutenegiambangatauho = B atau o = B/h = 9,9 x 10-19 J / 6,62 x 10-34 Js = 1,5 x1015 Hz

  6. Time Dilation, Time Comparisons • Observer O measures a longer time interval than observer O’

  7. The Twin Paradox – The Situation • A thought experiment involving a set of twins, Speedo and Goslo • Speedo travels to Planet X, 20 light years from earth • His ship travels at 0.95c • After reaching planet X, he immediately returns to earth at the same speed • When Speedo returns, he has aged 13 years, but Goslo has aged 42 years

  8. Length Contraction – Equation • Length contraction takes place only along the direction of motion

  9. Relativistic Momentum • To account for conservation of momentum in all inertial frames, the definition must be modified • v is the speed of the particle, m is its mass as measured by an observer at rest with respect to the mass • When v << c, the denominator approaches 1 and so p approaches mv

  10. Relativistic Energy • The definition of kinetic energy requires modification in relativistic mechanics • KE = mc2 – mc2 • The term mc2 is called the rest energy of the object and is independent of its speed • The term mc2 is the total energy, E, of the object and depends on its speed and its rest energy

More Related