Su'aalo Tusaale ah oo ku saabsan Doodda Fiisigiska Nukliyeerka iyo Raajada
Fiisigiska nukliyeerka iyo firfircoonida shucaaca waa laamo fiisigis oo ka shaqeeya daraasadda nuclei-ga atomiga iyo ifafaalaha burburka shucaaca ee ka dhaca nuclei-gan. Barashada fikradahan aasaasiga ah waa muhiim dhinacyo kala duwan, oo ay ku jiraan daawada, tamarta nukliyeerka, iyo sayniska agabka. Maqaalkan, waxaan ka hadli doonaa dhowr tusaale oo dhibaatooyin ah oo la xiriira fiisigiska nukliyeerka iyo firfircoonida shucaaca, iyo sidoo kale sharraxaaddooda, si ay kaaga caawiyaan fahamkaaga.
Hordhac Aasaasi ah oo ku saabsan Fiisigiska Nukliyeerka iyo Raajada
Kahor inta aynaan guda gelin su'aalaha tusaalaha ah, waa fikrad wanaagsan in dib loo eego qaar ka mid ah fikradaha aasaasiga ah:
– Nukliyeerka Atomiga: Waxay ka kooban tahay borotoonno iyo neutrons. Borotoonnadu waxay leeyihiin dallac togan, halka neutronnadu aanay lahayn dallac.
– Raajada: Habka burburka unugyada aan degganayn una beddelo nuclei aad u deggan iyadoo la sii daayo walxaha ama shucaaca.
– Noocyada Burburka Raajada: Alpha (\(\alpha\)), beta (\(\beta\)), iyo gamma (\(\gamma\)) burbur.
– Sharciga Burburka Raajada: Wuxuu qeexayaa sida tirada nuclei-yada shucaaca ay u yaraato waqti ka dib.
Su'aal Tusaale ah 1: Tamarta Cufnaanta iyo Burburka
Su'aal:
Nukliyeerka uranium-238 wuxuu isu beddelaa thorium-234 iyadoo la sii daayo walxaha alpha. Haddii cufka uranium-238 uu yahay 238.0508 u, cufka thorium-234 waa 234.0436 u, cufka walxaha alpha-na waa 4.0026 u, xisaabi tamarta lagu sii daayay burburkan.
Dood:
Tamarta lagu sii daayo habka qudhunka waxaa lagu xisaabin karaa iyadoo la adeegsanayo xiriirka ka dhexeeya cufka iyo tamarta oo lagu bixiyay isla'egta Einstein \(E=mc^2\).
1. Xisaabi cufka maqan:
\( \Delta m = (mass_{U-238}) – (mass_{Th-234} + mass_{\alpha}) \)
\( = 238.0508 – (234.0436 + 4.0026) \)
\( = 238.0508 – 238.0462 \)
\( = 0.0046\, u \)
2. Cufka lumay u beddel tamar adoo isticmaalaya \( c^2 \):
\( E = \Delta m \times 931.5\, MeV/u \)
\( = 0.0046 \ jeer 931.5 \)
\( \qiyaastii 4.29\, MeV \)
Markaa, tamarta lagu sii daayo burburkan waa qiyaastii 4.29 MeV.
Su'aal Tusaale 2: Nus-Nolol iyo Hawlo
Su'aal:
Muunad shucaac ah waxay marka hore leedahay dhaqdhaqaaq \(A_0 \) oo ah 1000 Bq. 10 saacadood ka dib, dhaqdhaqaaqeedu wuxuu hoos ugu dhacaa 125 Bq. Go'aami kala bar nolosha walaxda shucaaca.
Dood:
Dhaqdhaqaaqa (A) ee walaxda shucaaca waxay si toos ah ugu dhigantaa tirada nuclei-ga shucaaca (N). Sharciga suuska shucaaca wuxuu leeyahay:
\[ A(t) = A_0 e^{-\lambda t} \]
Halkee \( \lambda \) uu yahay joogtada qudhunka:
1. Xisaabi joogtada burburka (\( \lambda \)):
\[ \frac{A(t)}{A_0} = e^{-\lambda t} \]
\[ \frac{125}{1000} = e^{-\lambda \times 10} \]
\[ 0.125 = e^{-\lambda \times 10} \]
\[ \ln(0.125) = -\lambda \times 10 \]
\[ \lambda = -\frac{\ln(0.125)}{10} \]
2. Go'aami nolosha nuskeeda (\( T_{1/2} \)):
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
\[ \lambda = \frac{\ln(8)}{10} = \frac{2.079}{10} = 0.2079 \, saacad^{-1} \]
\[ T_{1/2} = \frac{\ln(2)}{0.2079} \qiyaastii 3.3 \, saacado \]
Nolosha badhkeed ee walaxda shucaaca waa qiyaastii 3.3 saacadood.
Tusaale 3: Duufka Beta iyo Antineutrinos
Su'aal:
Nucleus-ka Cobalt-60 wuxuu ku burburaa isagoo adeegsanaya beta-minus si uu u sameeyo nucleus Nickel-60 ah. Qor falgalka nukliyeerka ee burburkan oo aqoonso walxaha ku lug leh.
Dood:
Beta-minus suuska (\(\beta^-\)) wuxuu dhacaa marka neutron-ka ku jira xudunta atamka uu isu beddelo proton, halka uu soo saarayo elektaroono (\(\beta^-\)) iyo antineutrino (\(\bar{\nu}_e\)).
Falcelinta nukliyeerka ee burburkan waa:
\[ _{27}^{60}Co \rightarrow _{28}^{60}Ni^ + e^- + \bar{\nu}_e \]
Halkee:
– \( _{27}^{60}Co \) waa Cobalt-60.
– \( _{28}^{60}Ni^ \) waa Nickel-60 oo xaalad kacsan ku jira.
– \( e^- \) waa elektaroon (qaybta beta-laga jaray).
- \ ( \bar {\nu}_e \) waa antineutrino.
Nikkel-60 oo la sameeyay badanaa wuxuu ku jiraa xaalad kacsan wuxuuna badanaa sii daayaa tamar dheeraad ah oo ah qaabka shucaaca gamma (\(\gamma\)) si uu u gaaro xaaladda dhulka. Falcelinta oo dhammaystiran waxaa loo qori karaa sidan:
\[ _{27}^{60}Co \rightarrow _{28}^{60}Ni + e^- + \bar{\nu}_e + \gamma \]
Su'aal Tusaale ah 4: Qiyaasta Shucaaca
Su'aal:
Haddii isha shucaaca gamma oo leh dhaqdhaqaaq 2 Curies ah la dhigo masaafo 1 mitir u jirta shay oo shucaaca uu nuugo shaygu muddo 5 daqiiqo ah, xisaabi qiyaasta shucaaca ee uu shaygu helay rems-ka. U qaado in xaddiga shucaaca la daboolay uu yahay 0.5 rad Curie kasta daqiiqaddii iyo qodobka tayada shucaaca gamma uu yahay 1.
Dood:
1. Xisaabi qiyaasta rad:
\[ \text{Qiyaasta (rad)} = \text{Qaddarka shucaaca} \times \text{Hawlgalka} \times \text{Waqtiga (daqiiqado)} \]
\[ = 0.5 \, rad/(Ci \cdot min) \times 2 \, Ci \ times 5 \, min \]
\[ = 5 \, rad \]
2. Xisaabi qiyaasta bareegyada:
\[ \text{Qiyaasta (rem)} = \text{Qiyaasta (rad)} \times \text{Qiyaasta Tayada} \]
\[ = 5 \, rad \times 1 \, (gamma) \]
\[ = 5 \, rem \]
Qiyaasta shucaaca ee uu helo shaygu waa 5 rem.
Xiritaanka
Markaan baranno dhibaatooyinka tusaalaha ah ee kor ku xusan, waxaan rajeyneynaa inaan si qoto dheer u fahamno fikradaha fiisigiska nukliyeerka iyo firfircoonida shucaaca. Waa muhiim inaad si joogto ah u dhaqanto dhibaatooyinka la midka ah si aad u noqoto mid aad ugu xeel dheer fahamka iyo adeegsiga fikradaha fiisigiska nukliyeerka. Waxbarasho wanaagsan!
