Nuclei of radioactive element `A` are produced at rate `t^(2)` (where `t` is time) at any time `t`. The element `A` has decay constant `lambda`. Let `N` be the number of nuclei of element `A` at any time `t`. At time `t=t_(0), dN//dt` is minimum. The number of nuclei of element `A` at time `t=t_(0)` is
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Atomic weight of a radioactive element is `M_(W)` gm. Radioactivity of m gm of its mass is :- (`N_(A) = ` Avogardro number, `lambda` = decay constant)
Correct Answer - B `R = lambdaN = lambda[(m)/(M_(w)) N_(A)]`
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If `[(lambda^(2)-2lambda+1,lambda-2),(1-lambda^(2)+3lambda,1-lambda^(2))]=Alambda^(2)+Blambda+C`, where A, B and C are matrices then find matrices B a
We have `[(lambda^(2)-2lambda+1,lambda-2),(1-lambda^(2)+3lambda,1-lambda^(2))]=Alambda^(2)+Blambda+C` Putting `lambda=0`, we get `C=[(1,-2),(1,1)]` Putting `lambda=1`, we get `A+B+C=[(0, -1),(3,0)]` ...(1) Putting `lambda=-1`, we get `A-B+C=[(4,-3),(-3,0)]` Subtracting (2) from (1), we get `2B[(0, -1),(3,0)]-[(4,-3),(-3,0)]=[(-4,2),(6,0)]` `:. B=[(-2,...
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Nuclei of `.^(64)Cu` can decay be electron capture (probability 61%) or by `beta^(+)` decay (39%). Half life of `.^(64)Cu` is 12.7 hour. Find the part
Correct Answer - 20.8hr.
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A radioactive element is produced in nuclear reactor at a constant rate R(=numbers of nuclei per second). Its half-life is `T_(1//2)`. How much time,
Correct Answer - `T_(1//2)`,No
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A radioactive nucleus A decays into B with a decay constnat `lambda_(1)`. B is alos radioactive and it decays into C with a decay constant of `lambda_
Correct Answer - (a)`(dN_(2))/(dt)=-lambda_(2)N_(2)+lambda_(1)N_(1)^(0)e^(-lambda_(1)t)` (b) `(A_(2))/(A_(1))=(lambda_(2))/(lambda_(2)-lambda_(1))` (c) `((1)/(lambda_(2)-lambda_(1)))ln((lambda_(2))/(lambda_(1)))`
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If the equations `4x^(2)-x-1=0` and `3x^(2)+(lambda+mu)x+lambda-mu=0` have a root common, then the rational values of `lambda` and `mu` are
Correct Answer - C `(c )` Roots of `4x^(2)-x-1=0` are irrational. So, one root common implies both roots are common. `:. (4)/(3)=(-1)/(lambda+mu)=(-1)/(lambda-mu)=(-2)/(2lambda)=(0)/(2mu)` So, `lambda=(-3)/(4)`, `mu=0`
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Find all values of `lambda` for which the `(lambda-1)x+(3lambda+1)y+2lambdaz=0(lambda-1)x+(4lambda-2)y+(lambda+3)z=0 2x+(3lambda+1)y+3(lambda-1)z=0` p
Correct Answer - D `(d)` `|{:(lambda-1,3lamda+1,2lambda),(lambda-1,4lambda-2,lambda+3),(2,3lambda+1,3(lambda-1)):}|=0` `implieslambda=0` or `3` If `lambda=0`, the equations become `-x+y=0`, `-x-2y+3z=0` and `2x+y-3z=0`, `:. (x)/(6-3)=(y)/(6-3)=(z)/(-1+4)`
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In a reactor, an element X decays to a radioactive element Y, at a constant rate r atoms per second. Each decay reaction releases energy `E_(1)`. Half
Correct Answer - `[(i)" "eta r[E_(1)+E_(2)(1-e^((t log 2)/T))], (ii)" "eta r(E_(1)+E_(2))]`
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A number `N_(0)` of atoms of a radio active element are placed inside a closed volume. The radioactive decay constant for the nucleus of this element
Correct Answer - `[N_(0)e^(-lambda_(2)t).(lambda_(1)N_(0))/(lambda_(2))(t-e^(-lambda_(2)t))]`
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A spectroscopic instrument can resolve two nearly wavelength `lambda and lambda = Delta lambda if lambda//Delta lambda` is smaller than `8000` This is
Correct Answer - [38]
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