The half-life of a nuclide is 737 hrs
Let the initial mass of nuclide is N° = 100
The final mass of nuclide after 407 hours
= 100 - 38 = 62
The expression for decay constant is
λ = 1/t ln(N°/Nt)
Substitute the given values and calculate the decay constant as,
λ = 1/407hr ln(100/62)
λ = 9.4 x 10^-4/hr
Now, the half-life is calculated as,
t1/2 = 0.693/λ
= 0.693/9.4 x 10^-4/hr
t1/2 = 737 hrs
Hence, the half-life of a nuclide is 737 hrs
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Answer:
Explanation:
Momentum P = Mass x Velocity
M = 4.5kg
V = 7m/s
4.5kg x 7m/s
= 31.5xkgm/s
L=IW( Angular momentum) at stationary origin (0,0)
I = 1/2 x Mr^2
L = 1/2 x 4.5x 31.5
L = 70.8kgm/s
At stationary point, (0,0) No coordinate exist
300
Explanation:
100 x 3 =300 simple and easy