A) 1442
B) 1554
C) 1652
D) 1458
Correct Answer: D
Solution :
[d] \[N={{N}_{{}^\circ }}{{e}^{-\lambda t}}\] \[1800=2000{{e}^{-\lambda \times 2}}\] \[\frac{9}{10}={{e}^{-2\lambda }}\Rightarrow {{e}^{-\lambda }}={{\left[ \frac{9}{10} \right]}^{1/2}}\] No. of nuclei left after 6s \[N={{N}_{{}^\circ }}{{e}^{-\lambda T}}\] \[N=2000{{\left[ \frac{9}{10} \right]}^{1/2\times 6}}=2000\times {{\left[ \frac{9}{10} \right]}^{3}}\] \[N=2000\times \frac{729}{1000}=1458\]
You need to login to perform this action.
You will be redirected in
3 sec
