`U^(235) + n^(1)rarr` fission product + neutron + 3.2`xx 10^(-11)j`. The energy released , when 1g of `u^(235)` finally undergoes fission , is

Monjur Alahi

`U^(235) + n^(1)rarr` fission product + neutron + 3.2`xx 10^(-11)j`. The energy released , when 1g of `u^(235)` finally undergoes fission , is
A. `12.75xx10^(8)kj`
B. `18.60xx10^(9)kj`
C. `8.21xx10^(7)kj`
D. `6.55xx10^(6)kj`


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - C
235 g of u-235 contains `6.023xx10^(23)` atoms
1g u-235=`(6.023xx10^(23))/(235)atoms`
`therefore` Energy relaeased =`(3.2xx10^(-11)xx6.023xx10^(23))/(235)j`
`=8.21xx10^(10)j`
`8.21xx10^(7) kj`

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
`CH_(3)CH_(2)CH_(2)Cl overset(alc. KOH)rarr B overset(HBr)rarr C overset("Na//ether")rarr D` In the above reaction, the product D is

Correct Answer - C `CH_(3)CH_(2)CH_(2)Cl overset(alc. KOH)rarr CH_(3)CH=CH_(2) overset(HBr)rarr CH_(3)-underset(Br)underset("| ")("C ")H-CH_(3)overset(Na)underset("ether")rarr underset("2,3-Dimethylbutane")(CH_(3)-overset(CH_(3))overset("| ")("C ")H-overset(CH_(3))overset("| ")("C ")H-CH_(3))`

2 Answers 1 views
Complete the following : (1)`._(7)N^(14)+ ._(2)He^(4)rarr ._(8)O^(17) + ......" "` , (2)`._(92)U^(235)+ ._(0)n^(1)rarr ._(55)A^(142) +. _(37)B^(92)+ .

(1)`._(7)N^(14)+ ._(2)He^(4)rarr ._(8)O^(17) + ._(1)H^(1) " "` , (2)`._(92)U^(235)+ ._(0)n^(1)rarr ._(55)A^(142) +. _(37)B^(92)+ 2 ._(0)n^(1)` (3)`._(29)Cu^(53)rarr ._(28)Ni ^(53) + ._(+1)e^(0) ` ,

2 Answers 1 views
Fill in the blanks (a) `._(92)^(235)U+._(0)^(1)n rarr ._(52)^(137)A+ ._(40)^(97) B + __________. (b) `._(34)^(82) Se rarr 2 _(-1)^(0)e+__________`.

Correct Answer - (a) `2_(0)^(1)n` , (b) ` ._(36)^(82) Kr` `._(92)^(235)U+_(0)^(1)nrarr ._(52)^(137)A+_(40)^(97)B+X ._(0)n^(1)` (as the reaction is balanced with respect to nuclear charge, the missing particle must be neutral i.e. neutron)...

2 Answers 1 views
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`=...

2 Answers 1 views
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)`

2 Answers 1 views
What is the energy released by fassion of 1 g of `U^(235)`? (Assume 200 Me V energy is liberated on fission of 1 nucleus)

Correct Answer - A (a) Energy relased in one fission of `""_(92)^(235)U` nucleus `=200 meV ` Mass of uranium = 1 g we know that , 235 g of ` ""^(235)U`...

2 Answers 1 views
`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`

2 Answers 1 views
`""_(92)U""^(235)` nucleus absorbes a neutron and disintegrate into `""_(54)Xe""^(139),""_(38)Sr""^(94)`, and x neutrons x is

Correct Answer - A `_(92)U^(256)` nucleus absorbs a neutron and then disntegratein `_(54)Xe^(139),_(38)Sr^(94)` and X. Thus `_(92U^(235)+_(0)n^(1)rarr_(54)Xe^(139)+_(38)Sr^(94)+3_(0)n^(1))` Hence the product is 3 neutrons.

2 Answers 1 views
The reaction `Aoverset(k)rarr` Product, is zero order while the reaction `Boverset(k)rarr` Product, is first order reaction. For what intial concentra

Correct Answer - A For zero order reaction, x=kt `:. (a)/(2)=kxxt_(1//2),i.e.,t_(1//2)=(a)/(2k)" ".....(i)` For frist order reaction, `t_(1//2)=(log_(e)2)/(k)" ".....(ii)` From eqs. (i) and (ii), `(a)/(2k)=(log_(e)2)/(k)` `a=log_(e)4M`

2 Answers 1 views
A deuterium nucleus at rest absorbs a Gamma ray photon of energy 2.21 MeV and splits into a proton and a neutron. The neutron was found to be at rest

Correct Answer - 1.008968u

2 Answers 2 views
X