JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2003

A nucleus with $Z=92$ emits the following in a sequence: $$\alpha ,{\beta ^ - },{\beta ^ - },\alpha ,\alpha ,\alpha ,\alpha ,\alpha ,{\beta ^ - },{\beta ^ - },\alpha ,{\beta ^ + },{\beta ^ + },\alpha$$

Then $Z$ of the resulting nucleus is

A
$76$
B
$78$
C
$82$
D
$74$

Explanation

The number of $\alpha$ - particles released $=8$

Therefore the atomic number should decrease by $16$

The number of ${\beta ^ - }$ - particles released $=4$

Therefore the atomic number should increase by $4.$

Also the number of ${\beta ^ + }$ particles released is $2,$ which

should decrease the atomic number by $2.$

Therefore the final atomic number is

$92-16+4-2=78$
2

AIEEE 2003

A radioactive sample at any instant has its disintegration rate $5000$ disintegrations per minute. After $5$ minutes, the rate is $1250$ disintegrations per minute. Then, the decay constant (per minute) is
A
$0.4$ $ln2$
B
$0.2$ $ln2$
C
$0.1$ $ln2$
D
$0.8$ $ln2$

Explanation

$\lambda = {1 \over t}{\log _e}{{{A_0}} \over A}$

$= {1 \over 5}{\log _e}{{5000} \over {1250}}$

$= 0.2{\log _e}4$

$= 0.4{\log _e}2$
3

AIEEE 2003

When a ${U^{238}}$ nucleus originally at rest, decays by emitting an alpha particle having a speed $'u',$ the recoil speed of the residual nucleus is
A
${{4\mu } \over {238}}$
B
$- {{4\mu } \over {234}}$
C
${{4\mu } \over {234}}$
D
$- {{4\mu } \over {238}}$

Explanation

Here, conservation of linear momentum can be applied

$238 \times 0 = 4u + 234v$

$\therefore$ $v = - {4 \over {234}}u$

$\therefore$ speed $= |\overrightarrow v | = {4 \over {234}}u$
4

AIEEE 2003

Which of the following radiations has the least wavelength ?
A
$\gamma$ - rays
B
$\beta$ - rays
C
$\alpha$ - rays
D
$X$ - rays

Explanation

The electromagnetic spectrum is as follows

$\therefore$ $\gamma$-rays has least wavelength.