which a `U^(238)` nucleus original at rest , decay by emitting an alpha particle having a speed `u` , the recoil speed of the residual nucleus is
A. `(4u)/(238)`
B. `-(4u)/(234)`
C. `(4u)/(234)`
D. `-(4u)/(238)`
1 Answers
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A ultra violet light bulb, emitting 400 nm and infrared bulb emitting 700 nm wavelength radiation each are rated at 130 W. Then the ratio of the numbe
Correct Answer - A `n_(1)((hc)/(lambda_(1)))=n_(2)((hc)/(lambda_(2)))="power"=130W` `n_(1)*n_(2)` number of photons emitted per second by UV & IR sources `(n_(1))/(n_(2))=(lambda_(1))/(lambda_(2))=(400nm)/(700nm)=(4)/(7) =0.57`
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A nucleus with Z =92 emits the following in a sequence: `alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(
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The analysis of a uranium reveals that ratio of mole of `.^(206)Pb`and `.^(238) U` in sample is 0.2 . If effective decay constant of process `.^(238)U
Correct Answer - D In the rock `("Moles of ".^(206)Pb)/("Moles of " .^(238 )U)=0.2=(0.2)/(1)` `{:(,.^(238)U,rarr,.^(206)Pb,),("Initial moles",1.2 "mole*",,0,"(*calculate by moles of"),(,1"mole",,0.2 "mole","U left & moles of Pb formed) "):}` `" " t=(1)/(lambda)...
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A sample of `U^(238) ("half life" = 4.5 xx 10^(9)yr)` ore is found to contain `23.8 g" of " U^(238) ` and `20.6g ` of `Pb^(206)`. Calculate the age of
Correct Answer - [`4.5 xx 10^(9)` year] `U^(238) rarr Pb^(206)` `((20.6)/(206))=0.1 "mol present"` i.e. 0.1 mol of `U^(238)` decay `0.1 xx 238=(23.8)` decay Initial mass of `U^(238)=(23.8 +23.8)=47.6` 23.8 left and...
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A stationary radioactive nucleus of mass 210 units disintegrates into an alpha particle of mass 4 units and residual nucleus of mass 206 units. If the
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The kinetic energy of `alpha`-particle emitted in the `alpha`-decay of `""_(88)Ra^(266)` is [given, mass number of Ra = 222 u]
Correct Answer - D (d) the nuclear should be `""_(88)Ra^(226) to ""_(86)Rn ^(222)+ alpha ` Energy corresponding mass defect `Q =[{:("(mass number of Ra )"),("-(Mass number of Ru )"),("-m (4He )"):}]Uxx931.5Me...
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Assetion In `alpha`-decay atomic number of daughter nucleus reduces by 2 units from the parent nucleus. Reason An `alpha` particle carries four units
Correct Answer - A We know that `alpha` particle carries two units of positve charge and four units of mass In `alpha` decays charge number of parent nucleus decreses by 2...
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If `A=[alphabetagamma-alpha]` is such that `A^2=I` , then `1+alpha^2+betagamma=0` (b) `1-alpha^2+betagamma=0` (c) `1-alpha^2-betagamma=0` (d) `1+alpha
Correct Answer - A `A^(2) =I` `implies [{:(alpha,beta),(gamma,-alpha):}][{:(alpha,beta),(gamma,-alpha):}]=I` `implies[{:(alpha^(2)+betagamma,alphabeta-betaalpha),(gammaalpha -alphagamma,betagamma +alpha^(2)):}]=I` `implies [{:(alpha^(2) +beta gamma ,0),(0, alpha^(2) +betagamma ):}]=[{:(1,0),(0,1):}]` `alpha^(2) +beta gamma =1 implies 1-alpha^(2) -betagamma =0`
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If a variable takes the discrete values `alpha-4`, `alpha -(7)/(2), alpha-(5)/(2), alpha-2,alpha+(1)/(2), alpha-(1)/(2), alpha+5(alpha gt 0)`, then th
Correct Answer - A Arrange the data as follows : `alpha-(7)/(2), alpha-3, alpha-(5)/(2),alpha-2, alpha-(1)/(2),alpha+(1)/(2),alpha+4,alpha+5` Median `=(1)/(2)` [value of 4th item+value of 5th item] `therefore " Median"=(alpha-2+alpha-(1)/(2))/(2)=(2alpha-(5)/(2))/(2)=alpha-(5)/(4)`
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Prove that `tan alpha+2 tan 2 alpha+2^(2)tan^(2)alpha+..............+2^(n-1)tan2^(n-1)alpha+2^(@)cot2^(@)alpha=cotalpha`
We know `tan theta=cot theta-2 cot 2 theta`. Putting `theta=alpha,2alpha,2^(2)alpha,....................` in (i), we get `tan alpha=(cot alpha-2 cot 2alpha)` `2(tan2alpha)=2(cot 2 alpha-2 cot 2 ^(2)alpha)` `2^(2)(tan2^(2)alpha)=2^(2)(cot2^(2)alpha-2cot2^(3)alpha)` `2^(n-1)(tan2^(n-1)alpha)=2^(n-1)(cot2^(n-1)alpha-2cos2^(n)alpha)` Adding, `tan alpha+2tan...
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