1

### JEE Main 2019 (Online) 9th January Morning Slot

A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 $\times$ 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive materail B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :
A
5 days and 10 days
B
10 days and 40 days
C
20 days and 5 days
D
20 days and 10 days

## Explanation

Let number of nuclei present in material A is NA and in meterial B is NB According to question,

NA = 2NB

We know, activity (A) = $\lambda$N.

$\therefore$   $\lambda$ANA = 10 . . . . . . (1)

and $\lambda$B NB = 20 . . . . . . . (2)

$\therefore$   By dividing (1) by (2), we get

${{{\lambda _A}{N_A}} \over {{\lambda _B}{N_B}}}$ = ${1 \over 2}$

$\Rightarrow$   ${{{\lambda _A}} \over {{\lambda _B}}} \times 2$ = ${1 \over 2}$ [as NA = 2NB]

$\Rightarrow$    ${{{\lambda _A}} \over {{\lambda _B}}}$ = ${1 \over 4}$

We know,

T$_{{1 \over 2}}$  $\propto$ ${1 \over \lambda }$

$\therefore$   ${{{{\left( {{T_{{1 \over 2}}}} \right)}_A}} \over {{{\left( {{T_{{1 \over 2}}}} \right)}_B}}}$ = ${{{\lambda _B}} \over {{\lambda _A}}} = 4$

$\Rightarrow$   ${\left( {{T_{{1 \over 2}}}} \right)_A}$ = 4${\left( {{T_{{1 \over 2}}}} \right)_B}$

$\therefore$   By checking options you can see possible value of ${\left( {{T_{{1 \over 2}}}} \right)_A}$ = 20 days and ${\left( {{T_{{1 \over 2}}}} \right)_B}$ = 5 days
2

### JEE Main 2019 (Online) 9th January Evening Slot

At a given instant, say t = 0, two radioactive substance A and B have equal activities. the ratio ${{{R_B}} \over {{R_A}}}$ of their activities after time t itself decays with time t as e$-$3t. If the half-life of A is ln2, the half-life of B is :
A
4ln2
B
${{\ln 2} \over 2}$
C
${{\ln 2} \over 4}$
D
2ln2

## Explanation

We know,

Activity (R) = R0 e$-$$\lambda t Given that, at t = 0 RA = RB \Rightarrow R0A e-$$\lambda$Ax0 = R0B e$-$ $\lambda$Bx0

$\Rightarrow$  R0A = R0B

Given that at time t,

${{{R_B}} \over {{R_A}}} = {e^{^{ - 3t}}}$

$\Rightarrow$   ${{{R_{0B}}\,{e^{ - {\lambda _B}t}}} \over {{R_{0A}}\,{e^{ - {\lambda _A}t}}}}$ = e$-$3t

$\Rightarrow$  ${e^{\left( {{\lambda _A} - {\lambda _B}} \right)t}} = {e^{ - 3t}}$

$\Rightarrow$  ($\lambda$A $-$ $\lambda$B)t = $-$3t

$\Rightarrow$ $\lambda$A $-$ $\lambda$B = $-$ 3 . . . . . (1)

We know half life, ${t_{{1 \over 2}}} = {{{{\ln }^2}} \over \lambda }$

$\therefore$  Half life of A is, ${{t_{{1 \over 2}}} = {{{{\ln }^2}} \over {{\lambda _A}}}}$

$\therefore$  $\lambda$A = ${{\ln 2} \over {{t_{{1 \over 2}}}}}$

= ${{\ln 2} \over {\ln 2}}$

= 1

From equation (1)     we get,

$\lambda$A $-$ $\lambda$B = $-$ 3

$\Rightarrow$  1 $-$ $\lambda$B = $-$ 3

$\Rightarrow$  $\lambda$B = 4

$\therefore$  Half life of B is $\left( {{t_{{1 \over 2}}}} \right) = {{\ln 2} \over {{\lambda _B}}} = {{\ln 2} \over 4}$
3

### JEE Main 2019 (Online) 10th January Morning Slot

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to -
A
200
B
150
C
400
D
360

## Explanation

at t = 0, A0 = ${{dN} \over {dt}}$ = 1600C/s

at t = 8s, A = 100 C/s

${A \over {{A_0}}}$ = ${1 \over {16}}$ in 8 sec

Therefor half life is t1/2 = 2 sec

$\therefore$   Activity at t = 6 will be 1600${\left( {{1 \over 2}} \right)^3}$

= 200 C/s
4

### JEE Main 2019 (Online) 10th January Evening Slot

Consider the nuclear fission

Ne20 $\to$ 2He4 + C12

Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -
A
8.3 MeV energy will be released
B
energy of 11.9 MeV has to be supplied
C
energy of 12.4 MeV will be supplied
D
energy of 3.6 MeV will be released

## Explanation

Ne20      $\to$  2He4 + C12

Q – value, EB = (BE)react $-$ (BE)product

= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)

= 9.72 MeV