Tiêu đề 13. NUCLEI.pdf ✅

BUY COURSE FROM STORE TO GET INSTANT ANSWER AND EXPLANATION OF EACH QUESTION SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 13. 13. NUCLEI #QID# 34595 (1.) A sample contains 16 g of a radioactive material, the half life of which is two days. After 32 days, the amount of radioactive material left in the sample is (a.) Less than 1 mg (b.) 1 4 g (c.) 1 2 g (d.) 1g Ans: a Exp: (a) Remaining amount = 16 × ( 1 2 ) 32/2 = 16 × ( 1 2 ) 16 = ( 1 2 ) 12 < 1mg #QID# 34596 (2.) Neutron is a particle, which is (a.) Charged and has spin (b.) Charged and has no spin (c.) Charge less and has spin (d.) Charge less and has no spin Ans: c #QID# 34597 (3.) The ratio of half-life times of two elements A and B is TA TB . The ratio of respectively decay constants λA λB is (a.) TB TA (b.) TA TB
BUY COURSE FROM STORE TO GET INSTANT ANSWER AND EXPLANATION OF EACH QUESTION SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 (c.) TA+TB TA (d.) TA−TB TA Ans: a Exp: (a) Half-life of a radioactive element T = 0.693 λ or T ∝ 1 λ ∴ λA λB = TB TA #QID# 34598 (4.) In the following reaction the value of ′X′ is 7N 14+2 He 4 → X + 1H 1 (a.) 8N 17 (b.) 8O 17 (c.) 7O 16 (d.) 7N 16 Ans: b Exp: (b) 7N 14 + 2He 4 → 8O 17 + 1H 1 #QID# 34599 (5.) If N1 = N0e −λt1, then the number of atoms decayed during time interval from t1and t2 (t2 > t1 ) will be (a.) Nt1 = Nt2 = No[e −λt1 − e −λr2] (b.) Nt2 = Nt1 = No[e −λt2 − e −λt1] (c.) Nt2 − Nt1 = No[e λt2 − e −λt1] (d.) None of the above Ans: a Exp: (a) Nt1 = N0e –λt1 Nt2 = N0e −λt2 ∴ Nt1 − Nt2 = N0(e −λt2 − e −λt2)
BUY COURSE FROM STORE TO GET INSTANT ANSWER AND EXPLANATION OF EACH QUESTION SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 #QID# 34600 (6.) The possible quantum numbers for 3d electrons are (a.) n = 3, l = 1, ml = +1, ms = − 1 2 (b.) n = 3, l = 2, ml = +2, ms = − 1 2 (c.) n = 3, l = 1, ml = −1, ms = + 1 2 (d.) n = 3, l = 0, ml = +1, ms = − 1 2 Ans: b #QID# 34601 (7.) Calculate the energy released when three α − particles combined to from a 12C nucleus , the mass defect is (atomic mass of 2He 4 is 4.002603 u) (a.) 0.007809 u (b.) 0.002603 u (c.) 4.002603 u (d.) 0.5 u Ans: a Exp: (a) Mass defect ∆m = Total mass of α − particles − mass of 12C nucleus = 3 × 4.002603 − 12 = 12.007809 − 12 = 0.007809 unit #QID# 34602 (8.) In a hydrogen atom, which of the following electronic transitions would involve the maximum energy change (a.) From n = 2 to n = 1 (b.) From n = 3 to n = 1 (c.) From n = 4 to n = 2 (d.) From n = 3 to n = 2
BUY COURSE FROM STORE TO GET INSTANT ANSWER AND EXPLANATION OF EACH QUESTION SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 Ans: b Exp: (b) From diagram E1 = −13.6 − (−3.4) = −10.2eV E2 = −13.6 − (−1.51) = −12.09eV E3 = −1.51 − (−0.85) = −0.66eV E4 = −3.4 − (−0.85) = (−2.55)eV E3 is least, i. e., frequency is lowest #QID# 34603 (9.) The energy equivalent to 1 mg of matter in MeV is (a.) 56.25 × 1022 (b.) 56.25 × 1024 (c.) 56.25 × 1026 (d.) 56.25 × 1028 Ans: a Exp: (a) 1amu (or 1 u)=1.6605402 × 10−27 kg = 1.6 × 10−24 g Moreover 1 amu is equivalent to 931 MeV 0r 1.6 × 10−24 g is equivalent to 931 MeV ∴ 1g is equivalent to 931 1.6×10−24 MeV and 10−3 g is equivalent to 931 1.6×10−24 × 10−3MeV = 5.6 × 1023 MeV #QID# 34604 (10.) The mass defect in particular nuclear reaction if 0.3 g. The amount of energy liberated in kilowatt hour is (Velocity of light= 3 × 108 ms −1 ) (a.) 1.5 × 106 (b.) 2.5 × 106 (c.) 3 × 106 (d.) 7.5 × 106 Ans: d