Tauira o ngā Pātai Kōrero mō te hihi-X
He hangarau hou ngā hihi-X, e mōhiotia whānuitia ana ko ngā rontgens, i roto i te rongoā me te ahumahi. Mai i te kitenga i a Wilhelm Conrad Röntgen i te tau 1895, kua whakatuwherahia e ngā hihi-X he matapihi nui mō ngā kaipūtaiao me ngā tohunga hauora ki te tiro i te hanganga o roto o ngā mea me te kore e pakaru. Heoi, he uaua te mārama ki ō rātou mātāpono mahi me ngā tono. Nō reira, ka matapakihia e tēnei tuhinga ngā tauira raruraru e pā ana ki ngā hihi-X hei āwhina i te whakaatu i te whakamahinga o tēnei hangarau i roto i ngā horopaki rerekē.
Hītori me ngā Mātāpono Taketake o ngā Hiko-X
I mua i te neke atu ki ngā tauira raruraru, he mea nui kia mārama ki ngā kaupapa matua. Ko ngā hihi-X he momo irahiko hikohiko me te roangaru poto rawa, e āhei ai rātou ki te uru ki te whānuitanga o ngā rauemi. Koinei te mea e āhei ai ngā hihi-X ki te whakaatu i ngā mea totoka, pērā i ngā wheua i roto i te tinana tangata.
Ko te kaupapa matua o ngā hihi-X e ahu mai ana i te taunekeneke i waenga i ngā hihi me te matū. Ina haere ngā hihi-X mā roto i tētahi mea, ka mimitihia ētahi, ko ētahi ka tukuna. Ko ngā mea matotoru ake, pērā i te wheua, ka mimitihia he nui ake ngā hihi-X i ngā kiko ngohengohe ake, pērā i te uaua me te kiri. Mā tēnei rerekētanga ka hangaia he ahua e taea ai e tātou te kite i ngā hanganga o roto.
Tauira Pātai 1: Te Tatau i te Inenga o te Hiko-X
Pātai
Ka tirotirohia tētahi tūroro ki te hihi-X me te horopeta hihi-X o te 0.03 Gray (Gy). Mena ko te pūngao katoa e mimitihia ana e te tinana o te tūroro he 1.5 Joules (J), tatauhia te taumaha tinana o te tūroro e whakaatuhia ana ki ngā hihi-X.
Solution
Ka taea te tatau i te horopeta irahiko (D) mā te whakamahi i te tātai:
\[ D = \frac{E}{m} \]
Kei hea,
– Ko te horopeta i roto i te Gy ko \( D \),
– Ko te pūngao e mimitihia ana i roto i ngā Joule te \( E \),
– Ko te papatipu i roto i ngā kirokaramu (kg) ko \( m \).
Hei kimi i te papatipu tinana (\( m \)), ka taea e tātou te whakarerekē i te tātai i runga ake nei ki:
\[ m = \frac{E}{D} \]
Whakakapia ngā uara e mōhiotia ana:
\[ m = \frac{1.5 \text{ J}}{0.03 \text{ Gy}} \]
Kia maumahara ko te 1 Gy = 1 J/kg, kātahi:
\[ m = \frac{1.5}{0.03} \text{ kg} \]
\[ m = 50 \text{ kg} \]
Nō reira, ko te taumaha o te tinana o te tūroro e whakaaturia ana ki ngā hihi-X he 50 kg.
Tauira Pātai 2: Te Tāutu i ngā Kiko mā te Whakamahi i ngā Hiko-X
Pātai
I roto i te whakaahua hihi-X, he kanapa ake te āhua o te wheua i te kiko ngohengohe. Whakamāramahia he aha i puta ai tēnei i roto i te horopaki o te taunekeneke a ngā hihi-X me ngā momo kiko rerekē.
Solution
Ko te mahi a ngā hihi-X i runga i ngā rerekētanga o te mimiti i waenga i ngā momo kiko rerekē. He nui ake te kiato me te tau ngota o te wheua i ngā kiko ngohengohe pērā i te uaua. Nō reira, he pai ake te wheua ki te mimiti i ngā hihi-X.
– Wheua: Nā te mea he kikī, ā, kei roto ngā huānga he nui ngā tau ngota pērā i te konupūmā, he nui ngā hihi-X e mimitia ana e te wheua. Nā tēnei ka iti ake ngā hihi-X e tae atu ana ki te kiriata, ki te pūoko rānei, ka mārama ake ngā wāhi o te ahua.
– Kiko Ngāwari: He iti ake te kiato me te tau ngota o ngā kiko ngāwari pērā i te uaua me te kiri. Kāore e nui ngā hihi-X e mimitihia ana e rātou, nō reira he nui ake ngā hihi-X e tukuna ana ki te kiriata, ki te pūoko rānei. Nā tēnei ka pōuri ake te āhua o ngā kiko ngāwari i roto i te ahua.
Mā ēnei rerekētanga o te tae, o te kanapa rānei ka taea e te tākuta, e te tohunga hangarau rānei te tautuhi ngāwari i ngā hanganga o roto i te tinana.
Tauira 3: Te Tātai i te Roanga Ngaru o ngā Hihi-X
Pātai
Mena he 124 keV (kilo-electronvolts) te kaha o te hihi-X, tatauhia tōna roangaru. Whakamahia te tātai:
\[ E = \frac{hc}{\lambda} \]
Kei hea,
– Ko te pūngao (i roto i ngā joule) ko \( E \),
– Ko te pūmau a Planck te \( h \), arā, ko te \( 6.626 \times 10^{-34} \) J·s (joule-hēkona),
– Ko te tere o te mārama te \( c \) , arā, \( 3 \times 10^8 \) m/s (mita ia hekona),
– Ko te \( \lambda \) te roanga ngaru (i roto i ngā mita).
Solution
Tuatahi, tahurihia te pūngao mai i te keV ki ngā joule. 1 eV (electronvolt) = \( 1.602 \times 10^{-19} \) J, kātahi:
\[ E = 124 \times 10^3 \text{ eV} = 124 \times 10^3 \times 1.602 \times 10^{-19} \text{ J} \]
\[ E = 1.985 \times 10^{-14} \text{ J} \]
Mā te whakamahi i te tātai:
\[ \lambda = \frac{hc}{E} \]
Whakakapia ngā uara e mōhiotia ana:
\[ \lambda = \frac{6.626 \times 10^{-34} \text{ J·s} \times 3 \times 10^8 \text{ m/s}}{1.985 \times 10^{-14} \text{ J}} \]
Tātaitanga:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{1.985 \times 10^{-14}} \text{ m} \]
\[ \lambda \tata ki te 1.002 \times 10^{-11} \text{ m} \]
Nō reira, ko te roanga ngaru o te hihi-X me te pūngao o te 124 keV he tata ki te \( 1.002 \times 10^{-11} \) mita, 0.1002 nm (nanometers) rānei.
Tauira Pātai 4: Te Whakamahinga o ngā hihi-X i roto i te Rongoā
Pātai
Kua tāpaetia he whati ringa o tētahi tūroro, ā, me tango he hihi-X. Mēnā ka whakamahia he hihi-X me te ngaohiko ngongo o te 100 kV me te ia o te 10 mA mō te 0.1 hēkona, e hia te tapeke o te utu e rere ana i roto i te ngongo?
Solution
Ka taea te tatau i te nui o te utu (Q) mā te whakamahi i te tātai:
\[ Q = I \whakareatia ki te t \]
Kei hea,
– Ko te utunga katoa i roto i ngā coulomb (C) ko \( Q \),
– Ko te iahiko i roto i ngā amperes (A) ko \( I \),
– Ko te \( t \) te wā i roto i ngā hēkona (s).
Whakakapia ngā uara e mōhiotia ana:
\[ I = 10 \kuputuhi{ mA} = 10 \times 10^{-3} \kuputuhi{ A} = 0.01 \kuputuhi{ A} \]
\[ t = 0.1 \text{ s} \]
\[ Q = 0.01 \kuputuhi{ A} \times 0.1 \kuputuhi{ s} \]
\[ Q = 0.001 \kuputuhi{ C} \]
Nō reira, ko te tapeke o te utu e rere ana i roto i te ngongo he 0.001 coulomb.
Whakamutunga
E whakaatu ana te tauira raruraru i runga ake nei i ētahi āhuatanga hangarau e pā ana ki te whakamahinga o ngā hihi-X, tae atu ki te tatau i te horopeta, te roanga ngaru, me te utu katoa. Nō reira, ka taea te whakamahi i te mārama ki ngā ariā taketake me ngā mātāpono mahi o ngā hihi-X ki te whakaoti rapanga uaua ake, motuhake ake. He mea nui te mōhio ki tēnei rauemi mō tētahi kaimahi e mahi ana i roto i te irahiko, te rongoā, te miihini karihi rānei kia pai ai te whakamahinga haumaru me te whai hua o ngā hihi-X.
