 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2003

The wavelengths involved in the spectrum of deuterium $\left( {{}_1^2\,D} \right)$ are slightly different from that of hydrogen spectrum, because
A
the size of the two nuclei are different
B
the nuclear forces are different in the two cases
C
the masses of the two nuclei are different
D
the attraction between the electron and the nucleus is different in the two cases

Explanation

The wavelength of spectrum is given by

${1 \over \lambda } = R{z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

where $R = {{1.097 \times {{10}^7}} \over {1 + {m \over M}}}$

where $m=$ mass of electron

$M=$ mass of nucleus.

For different $M,R$ is different and therefore $\lambda$ is different
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

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

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

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