Title 13. NUCLEI.pdf ✅

#QID# 5974 (1.)  − particle consists of (a.) 2 protons only (b.) 2 protons and 2 neutrons only (c.) 2 electrons, 2 protons and 2 neutrons (d.) 2 electrons and 4 protons only(NEET 2019) Ans: b Exp: (b): Alpha particle is a positively charged particle. It is identical to the nucleus of the helium ( ) 4 2He atom, so it contains 2 protons and 2 neutrons. #QID# 5976 (2.) The rate of radioactive disintegration at an instant for a radioactive sample of half life 9 2.2 10  s is 10 1 10 s − . The number of radioactive atoms in the sample at that instant is, (a.) 20 3.17 10  (b.) 17 3.17 10  (c.) 18 3.17 10  (d.) 19 3.17 10  (Odisha NEET 2019) Ans: d Exp: (d): Given, 9 1/2 t s =  2.2 10 and rate of radioactive disintegration, 10 1 10 dN s dt − = 10 1 9 1/2 0.693 0.693 3.15 10 2.2 10 s t  − − = = =   Now, we know that, 0 t N N e− =  0 dN t N e N dt    − = − = −  10 10 19 10 3.15 10 3.17 10 N N − =    =  #QID# 5977 (3.) For a radioactive material, half‐life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is (a.) 20 (b.) 10 (c.) 30 (d.) 15 (NEET 2018) Ans: a Exp: (a) : Number of nuclei remaining, N = − = 600 450 150 According to the law of radioactive decay, 1/2 0 1 2   =     t N T N ; where N0 is the number of nuclei initially.

Exp: (d): N0 = Nuclei at time t = 0 N1 = Remaining nuclei after 40% decay ( ) 0 0 = − = 1 0.4 0.6 N N N2 = Remaining nuclei after 85% decay ( ) 0 0 = − = 1 0.85 0.15 N N 2 0 2 1 0 0.15 1 1( ) 0.6 4 2 N N N N  = = = Hence, two half life is required between 40% decay and 85% decay of a radioactive substance. Time taken 1/2 = =  = 2 2 30 min 60 min t #QID# 5985 (6.) A nucleus of uranium decays at rest into nuclei of thorium and helium. Then (a.) The helium nucleus has more momentum than the thorium nucleus. (b.) The helium nucleus has less kinetic energy than the thorium nucleus. (c.) The helium nucleus has more kinetic energy than the thorium nucleus. (d.) The helium nucleus has less momentum than the thorium nucleus. (2015) Ans: c Exp: (c) : If p and Th p and He p are the momenta of thorium and helium nuclei respectively, then according to law of conservation of linear momentum 0 Th He = + p p or Th He p = − p ‐ve sign shows that both are moving in opposite directions. But in magnitude Th He p p = If mTh and mHe are the masses of thorium and helium nuclei respectively, then Kinetic energy of thorium nucleus is 2 2 Th Th Th p K m = and that of helium nucleus is KHe = 2 2 He He p m Th He K K  = 2 Th He He Th p m p m             But Th He p p = and m m He Th    K K Th He or K K He Th  Thus the helium nucleus has more kinetic energy than the thorium nucleus. #QID# 5988 (7.) The binding energy per nucleon of 3 7 Li and 2 4He nuclei are 5.60 MeV and 7.06 MeV respectively In the nuclear reaction 7 1 4 4 3 1 2 2 Li H He He Q + → + + the value of energy Q released is (a.) 19.6MeV (b.) −2.4MeV (c.) 8.4MeV (d.) 17.3MeV (2014) Ans: d
Exp: (d) : Binding energy of 3 7 Li nucleus =  = 7 5.60 39.2 MeV MeV Binding energy of 2 4He nucleus =  = 4 7.06 28.24 MeV MeV The reaction is 7 4 1 3 1 2 Li H He Q + → + 2( )  = Q 2 (BE of 2 4He ) − ( 3 ) 7 BE Li =  − 2 28.24 39.2 MeV MeV = − = 56.48 39.2 17.28 MeV MeV MeV #QID# 5989 (8.) A radioisotope X with a half-life 9 1.4 10  years decays to Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1 : 7. The age of the rock is (a.) 9 1.96 10  years (b.) 9 3.92 10  years (c.) 9 4.20 10  years (d.) 9 8.40 10  years (2014) Ans: c Exp: (c): X Y → Number of nuclei at t = 0 N0 0 Number of nuclei after time t N x 0 − x (As per question) 0 1 7 N x x − = 0 7 7 N x x − = or 0 7 8 x N =  Remaining nuclei of isotope X 3 0 0 0 0 0 7 1 1 8 8 2 N x N N N N   = − = − = =     So three halflives would have been passed. 9 1/2  = =   t nT 3 1.4 10 years 9 =  4.2 10 years Hence, the age of the rock is 9 4.2 10  years. #QID# 5991 (9.) A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u . The energy liberated per u is ( given 1 u MeV = 931 ) (a.) 6.675 MeV (b.) 13.35 MeV