1

### 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
2

### 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
3

### JEE Main 2019 (Online) 11th January Morning Slot

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980$\mathop A\limits^ \circ$. The radius of the atom in the excited state, in terms of Bohr radius a0 will be : (hc = 12500 eV$\mathop A\limits^ \circ$)
A
4a0
B
9a0
C
25a0
D
16a0

## Explanation

Energy of photon $= {{12500} \over {980}} = 12.75eV$

$\therefore$  Electron will excite to n = 4

Since 'R' $\propto$ n2

$\therefore$  Radius of atom will be 16a0
4

### JEE Main 2019 (Online) 11th January Evening Slot

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:
A
${{25} \over {16}}$ $\lambda$
B
${{27} \over {20}}$ $\lambda$
C
${{16} \over {25}}$ $\lambda$
D
${{20} \over {27}}$ $\lambda$

## Explanation

For M $\to$ L steel

${1 \over \lambda }$ = K $\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}$

for N $\to$ L

${1 \over {\lambda '}}$ = K$\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}$

$\lambda ' = {{20} \over {27}}\lambda$