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True or False Statements : In the nuclear reaction `X^(200) rarr A ^(110)+B^(90)`. If the binding energy per nucleon for X, A and B is 7.4 MeV, 8.2.Me

Monjur Alahi

True or False Statements :
In the nuclear reaction `X^(200) rarr A ^(110)+B^(90)`. If the binding energy per nucleon for X, A and B is 7.4 MeV, 8.2.MeV and 8.2 MeV respectively, the energy released is 160 MeV.


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Salim Ahmed Kawsar

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Correct Answer - [T]
`DeltaE=E(B.E.)_(P)-E(B.E.)_(R)`
`rArr (110xx8.2+90xx8.2)-(200xx7.4)=160` MeV

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If the binding energy per nucleon in `._(3)Li^(7)` and `._(2)He^(4)` nuclei are respectively `5.60` MeV and `7.06` MeV, then the ebergy of proton in t

Correct Answer - B In a nuclear reaction the energy remains conservest `P + ._(3)^(7)Li rarr 2(._(2)^(4)He)` energy of proton `+ 7(5.60) = 2(4 xx 7.06)` `rArr` energy of `= 17.28...

2 Answers 1 views
What will be the energy change in the following nuclear reaction, `X^(40) + ._(0)n^(1) rarr Y^(30)+Z^(11)` if binding energy per nucleon of X,Y and Z

Correct Answer - B Since B.E./Nucleon of Reactants is high energy will be absorbed, Net energy absorbed =(Bond energy of Reactant nucleus )-(Total bond energy of Product Nuclei) ` =(40 xx...

2 Answers 1 views
Fill in the blanks with appropriate items : This binding energies per nucleon for `._(1)H^(2) ` and `._(2)H^(4)` are ` 1.1MeV and 7.0 MeV` respectivel

Correct Answer - `23.6 MeV` `2 ._(1)H^(2) rarr ._(2)He^(4)` `Delta E=rArr(2+2)(7)-2(1+1)(1.1)` `rArr 23.6 ` MeV

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Calculate the energy released in joules and MeV in the following nuclear reaction : `._(1)^(2) H+_(1)^(2) rarr _(2)^(3)He + _(0)^(1)n` Assume that the

Correct Answer - `[Delta E =2.28 xx 10^(-12)J =14.25 MeV ]` `DeltaE=Delta m xx931.5 Me rArrDeltam=SigmaM_(R)-SigmaM_(P) rArr Deltam=2(2.020)-(3.0160+1.0087)` `rArrDeltaE=Deltamxx931.5=14.25 MeV`

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In a nuclear reactor, the number of `U^(235)` nuclei undergoing fissions per second is `4xx10^(20).` If the energy releases per fission is 250 MeV, th

Correct Answer - B (B) the fission per second `=4xx10^(20)` So , the energy relased per second `=4xx10^(20) xx250MeV ` `=4xx10^(20) xx250xx10^(6) xx1.6xx10^(-19) J` therefore , energy relased in 10 h`=...

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The energy released by the fission of one uranium atom is 200 MeV. The number of fission per second required to prodice 6.4W power is

Correct Answer - A (a) Let there is n number of fission per second produces a power of 6.4 W, then `nxx200x10^(6) xx16x10^(-19) Js^(-1) =6.4Js^(-1)` `therefore n=(6.4)/( 200xx10^(-13) xx1.6)=(4)/(2xx10^(-11))=2xx10^(11)`

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The binding energy per nucleon of `._(3)^(7) Li` and `._(2)^(4)He` nuclei are `5.60` MeV and `7.06` MeV, respectively. In the nuclear reaction `._(3)^

Correct Answer - D The binding energy for `""_(1)H^(1)` is around zero and also not given in the question so we can igonor it `Q=2(4xx7.06)-7xx(5.60)=(8xx7.06)-(7xx5.60)` `=56.48-39.2=17.28MeV~=17.3 MeV`

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`A U^(235)` reactor generated power at a rate of P producting `2xx10^(18)` fussion per second. The energy released per fission is 185 MeV. The value o

Correct Answer - A Power, `P=(nE)/(t)` Given, `n=2xx10^(18)` fission per second E=185MeV Here, `P=(2xx10^(18)xx185xx10^(6)eV)/(1)` `implies=2xx10^(18)xx185xx10^(6)xx16xx10^(-19)=59.2MW`

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It is proposed to use the nuclear fusion reaction, `._1^2H+_1^2Hrarr_2^4He` in a nuclear reactor `200 MW` rating. If the energy from the above reactio

Energy released in the nuclear fusion is `Q = Deltamc^(2) = Deltam(931) MeV` (where `Deltam` is in amu) `Q = (2 xx 2.0141 - 4.0026) xx 931 MeV = 23.834...

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The binding energy of deuteron is `2.2` MeV and that of `._(2)^(4)He` is `28` MeV. If two deuterons are fused to form one `._(2)^(4)He`, th `n` the en

Correct Answer - A `2 "Deuteron rarr ._(2)^(4)He` `Q = "BE of Product" - "BE of reactant"` `= [28 MeV - 2(2.2 MeV)]` `= [28-4.4 ] = 23.6 MeV`

2 Answers 1 views

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