Science without religion is lame, religion without science is blind.

– Albert Einstein

KEAM Engineering 2009 Exam will begin on 25^{th} of this

(1) The nuclear radius of a certain nucleus is 7.2 fm and it has a charge of 1.28×10^{–17}C.The number of neutrons inside the nucleus is

(a) 136

(b) 142

(c) 140

(d) 132

(e) 126

The nuclear radius *R* is related to the mass number *A* as

*R = *1.2 *A*^{1/3}

On substituting for *R* in this equation we obtain *A*^{1/3} = 6 so that *A =* 216.

Since the total charge of the nucleus (carried by all the protons) is given as 1.28×10^{–17} coulomb and the charge on a proton is 1.6×10^{–19} coulomb, the number of protons (*Z*) in the nucleus is (1.28×10^{–17}) /(1.6×10^{–19}) = 80.

Therefore, the number of neutrons in the nucleus, *N = A *–* Z* = 216 – 80 = 136.

(2) The energy released in the fission of 1 kg of _{92}U^{235} is (Energy per fission = 200 MeV)

(a) 5.1×10^{26 }eV

(b) 5.1×10^{26 }J

(c) 8.2×10^{13 }J

(e)** **5.1×10^{23 }MeV** **

1 kg of _{92}U^{235} contains 1/0.235 moles and each mole contains 6.02×10^{23} atoms. Therefore1 kg of _{92}U^{235} contains (6.02×10^{23})/ 0.235 atoms.

Therefore, the energy *E *released in the fission of 1 kg of _{92}U^{235} is given by

*E = *(6.02×10^{23}) ×200 / 0.235 MeV = 5.1×10^{26 }MeV = 5.1×10^{32 }eV

Since 1 eV = 1.6×10^{–19 }joule, the above energy is (5.1×10 ^{32})×(1.6×10^{–19})^{ }joule. This works out to approximately **8.2****×10 ^{13 }J.**

(3) Which one of the following statements is true if half life of a radioactive substance is 1 month?

(a) 7/8^{th} part of the substance will disintegrate in 3 months

(b) 1/8^{th} part of the substance will remain undecayed at the end of 4 months

(c) The substance will disintegrate completely in 4 months

(d) 1/16^{th} part of the substance will remain undecayed at the end of 3 months

(e)** **The substance will disintegrate completely in 2 months

The number of nuclei* N* remaining *undecayed *at the end of *n* half lives is given by

*N = N*_{0}/2^{n} where *N*_{0} is the initial number.

Therefore, at the end of 3 *N*_{0}/2^{3} = *N*_{0}/8.

This means that 7/8^{th} part of the substance will disintegrate in 3 months [Option (a)].

You will find a few more multiple choice questions (with solution) on nuclear physics here.

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