The half-life of a radioactive substance is  30 minutes. The time (in minute) taken between 40% decay and 85% decay of the same radioactive substance is

(1) 15           

(2) 30

(3) 45           

(4) 60

If the radius of the ₁₃Al²⁷ nucleus is taken to R_Al then the radius of ₅₃Te¹²⁵ nucleus is nearly

1. (53/13)^1/3 R_Al

2. 5/3 R_Al

3. 3/5 R_Al

4. (13/53)^1/3 R_Al

The binding energy per nucleon of  ${}_3{}^7\mathrm{Li}$and ${}_2{}^4\mathrm{He}$nuclei are 5.60meV and 7.06meV, respectively. 
In the nuclear reaction ${}_3{}^7\mathrm{Li} + {}_1{}^1\mathrm{H} \to {}_2{}^4\mathrm{He} + {}_2{}^4\mathrm{He} + \mathrm{Q}$  , the value of energy Q released is -

(a)19.6MeV

(b)-2.4MeV

(c)8.4MeV

(d)17.3MeV

If the nuclear radius of ${}_{27}Al$ is 3.6 Fermi, the approximate nuclear radius of ${}_{64}Cu$ in Fermi is

(a)2.4                                         (b)1.2

(c)4.8                                         (d)3.6

The half-life of a radioactive nucleus is 50 days. The time interval $({\mathrm{t}}_2-{\mathrm{t}}_1)$ between the time ${\mathrm{t}}_2$ when $\frac{2}{3}$ of it has decayed and the time ${\mathrm{t}}_1$ when $\frac{1}{3}$ of it had decayed is:

1. 30 days

2. 50 days

3. 60 days

4. 15 days

Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t=0, the number of P species are $4N_0$ and that of Q is $N_0.$Half-life of P(for conversion to R) is 1 min whereas that of Q is 2 min. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be: 

1. $3N_0$                                         

2. $\frac{9N_0}{2}$

3. $\frac{5N_0}{2}$                                       

4. $2N_0$

The activity of a radioactive sample is measured as $N_0$ counts per minute at t=0 and $N_0/e$ counts per minute at t=5 min.The time (in minute) at which the activity reduces to half its value is 

(1) ${\log }_{e}2/5$                                                         

(2) $\frac{5}{{\log }_e 2}$

(3) $5 {\log }_{10 }2$                                                         

(4) $5 {\log }_e 2$

The decay constant of a radio isotope is $\mathrm{\lambda }.$ If ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$ are its activities at times ${\mathrm{t}}_1$ and ${\mathrm{t}}_2$ respectively, the number of nuclei which have decayed during the time $({\mathrm{t}}_1-{\mathrm{t}}_2)\; is:$

1. ${\mathrm{A}}_1{\mathrm{t}}_1-{\mathrm{A}}_2{\mathrm{t}}_2$                                       

2. ${\mathrm{A}}_1-{\mathrm{A}}_2$

3. $\frac{({\mathrm{A}}_1-{\mathrm{A}}_2)}{\mathrm{\lambda }}$                                         

4. $\mathrm{\lambda }({\mathrm{A}}_1-{\mathrm{A}}_2)$ 

Two nuclei have their mass numbers in the ratio of 1:3. The ratio of their nuclear densities would be 

(1) 1:3

(2) 3:1

(3) ${\left(3\right)}^{1/3}:1$

(4) 1:1

Radioactive material 'A' has decay constant '8\(\lambda\)' and material 'B' has a decay constant '\(\lambda\)'. Initially, they have the same number of nuclei. After what time, the ratio of the number of nuclei of material 'A' to that of 'B' will be \(\frac{1}{e}\)?

\(1 .   \frac{1}{7 \lambda}\)
\(2 .   \frac{1}{8 \lambda}\)
\(3 .   \frac{1}{9 \lambda}\)
\(4 .   \frac{1}{\lambda}\)