Makalu

Physics: Energy quantization

  • A darussanmu na ilmin kimiyyar lissafi wato (physics) yau zamu yi karatunmu ne akan wani maudu’i da naga dalibai da yawa na wahala da shi a Jami'a. Hakan ya samo asali ne sakamakon rashin samun foundation dinsa tun daga sakandire, idan dalibi ya fahimce shi daga sakandire in yaje jami’a ya kan zame mishi da sauki a maimakon ganinsa kamar wani sabon abu.

    Mene ne energy quantization?

    Energy quantization dai kamar yadda ya zo daga quantum theory wanda Max Planck ya yi suggesting ya ke cewa, heat ko kuma electromagnetic radiation is emitted in fixed, discrete or separate amounts known as quanta. Yanzu zamu sansu a fomula da kuma bayanan kowannansu. The energy E of a photon or quantum is given by, E = hƒ, E = hc/.

    Yanzu zamu dauki kowanansu musan fassararshi da kuma unit dinsa tare da yawansa a cikin lissafi. Where

    h = plancks constant = 6.6×10-34Js

    c = velocity of electromagnetic waves in free space = 3.0×108ms-1

    ʎ = wavelength of electromagnetic wave (m)

    ƒ = frequency of electromagnetic wave (Hz)

    Electrons in an atom have definite energy levels with definite values like Eo, E1, E2, E3………etc. Energy na electrons ya na iya canzawa daga wani level din zuwa wani. Kamar misali, energy na electrons din zai iya karuwa daga lowest energy level ko kuma ground state kamar haka, Eo zuwa E1 ko kuma ya dinga raguwa daga E3 zuwa lower energy level, kamar misalin daga E3 ya dinga raguwa zuwa E2 ko kuma E1.

    Bayan energy level akwai kuma abinda ake ce ma energy change wanda shima ana amfani da fomula ne wajan lissafin energy level a yayin da suke canzawa gashinan zamu gansu kamar haka; Energy Change, E = En – Eo = hƒ ko kuma yana iya zama E = En – Eo = hc / ʎ

    Where En = Energy in excited state, n = 1,2,3…..

    Eo = Energy in ground state.

    Ku duba: Darasi game da method of mixtures

    The greater the energy change, the greater will be the frequency of the emitted radiate on. Tunda duk mun san mene ne suke nufi kuma munga fomulolinsu yanzu zamu dauki misalai mu amsa wanda tambayoyine da suke fitowa a jarabawar aji shida ta gama sakandire wato (WAEC, NECO, JAMB).

    Misali na daya:

    An electron makes a transition from a certain energy level Ek to the ground state Eo. If the frequency of emission is 8.0×1014Hz, the energy emitted is. A.8.25×10-19 J B.5.28×10-19J C. 8.25×1019 J JAMB 2003/8 ( h = 6.6×10-34 Js)

    Amsar tambaya:

    Idan muka duba tambayar zamu ga an bamu frequency sannan ga constant an bamu shima agefe wato planck’s constant amma ya na da kyau mu rike shi koda tambaya tazo ba kuma akawo mana shiba.

    Ga data kamar haka:

    Frequency ƒ = 8.0×1014Hz,

    h = 6.6×10-34 Js 

    Yanzu zamu dauko fomula musa kamar haka:

    Energy emitted E = hƒ E = 6.6×10-34 8.0×1014 = 5.28×10-19J

    Misali na biyu:

    Calculate the energy carried by an X-ray of wavelength 6.0×10-10 m, [planck’s constant = 6.6×10-34 Js, velocity of light = 3.0×108ms-] WAEC 1998/2005

    Amsar tambaya:

    Wave length, ʎ = 6.0×10-10, velocity of light = 3.0×108ms-1 h = 6.6×10-34 Js,  yanzu zamu dauki fomula sai musa abubuwan da aka bamu, Energy E = hƒ or E = hc / ʎ E = 6.6×10-34 × 3.0×108 / 6.0×10-10 = 3.3×10-16 J 

    Misali na uku

    An electron makes a transition from -3.4eV energy level to -13.6eV energy level. Calculate the Value of the loss of energy due to transition in joules. Frequency of the emitted radiation [ 1eV = 1.610-19J; h = 6.6×10-34Js] NECO 2003/5

    Amsar tambaya:

    Data

    Initial energy level Ei = -3.4eV;

    final energy level Eƒ = -13.6eV,

    dama an riga an bamu wadansu constant din gasu nan kamar haka:

    1eV = 1.610-19J, 

    h = 6.6×10-34Js

    Yanzu tunda mun fitar da abubuwan da aka bamu a tambayar, sai mu dauki tambayoyin da ake so mu amsa guda biyu kowannensu mu amsa su daban daban zamu fara da value of the loss energy due to transition in joules wanda in ana son lissafinsa ana amfani ne da wannan fomula, gata

    Loss of energy, E = Ei - Eƒ = -3.4 (13.6) = 3.4 13.6 = 10.2eV =10.2 1.6 10-19 = 1.63 10-18J From E = hƒ, frequency ƒ = E / h = 1.63 10-18 / 6.610-34 = 2.5 1015 Hz

    Ku duba wannan makala da ta yi bayani akan yadda ake measurement of heat capacity

    Misali na hudu:

    An atom radiates 1.510-19J when an electron jumps from one level to another. What is the wavelength of the emitted radiation? [planck’s constant = 6.6×10-34Js] [ speed of light in a vacuum = 3.0×108ms-1] NECO 2005/8

    Amsar tambaya:

    Data

    Energy radiated, E = 1.5×10-19J,

    h = 6.6×10-34,

    c = 3.0×108ms-1

    Yanzu zamu dauki fomula mu yi amfani da abubuwanda aka bamu a tambayar, 

    E = hc/ʎ wavelenght, ʎ = h / E = 6.6×10-34 3.0×108 / 1.510-19 = 1.3210-6 m

    Misali na biyar:

    In a model of hydrogen atom, the energy levels Wn are given by the formula Wn = R/n2 where n is an integer and R is a constant. Determine the energy released in the transition from n = 3 to n = 2. WAEC 2002/4/5

    Amsar Tambaya:

    Energy released E = W3 W2 W3 = R / (3)2 = R/9, W2 = R / (2)2 = R / 4 W3 W2 = = - + = = 9R 4R / 36 =

    Wadannan sune misalan da muka kawo muku a karkashin wannan maudu’i. Idan dalibai suka karanta sosai kuma suka bi yadda muka amsa tambayoyin za su fahimci karatun sosai sannan kuma ku yi kokarin kara bincike a wasu textbooks din na physics domin samun cikakken bayan da fahimtar karatun. Ku ci gaba da bibiyarmu don samun c igaban wannan maudu’i saboda wanan na farkon kamar foundation muka yi akwai maudu’ai masu zuwa kamar haka, Electron volt and kinetic energy,Photo electric Effect; Einstein Equation,stopping potential,wave-particle duality da kuma Heisenberg Uncertainty Principle. wadannan sune topics masu zuwa insha Allah fatan za ku ci gaba da bibiyarmu, Allah karo ilmi da fahimta mai amfani ameen.

    Kuna iya karanta: Darasi game da resistivity and conductivity

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