 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2003

In the nuclear fusion reaction $${}_1^2H + {}_1^3H \to {}_2^4He + n$$
given that the repulsive potential energy between the two nuclei is $\sim 7.7 \times {10^{ - 14}}J$, the temperature at which the gases must be heated to initiate the reaction is nearly
[ Boltzmann's Constant $k = 1.38 \times {10^{ - 23}}\,J/K$ ]
A
${10^7}\,\,K$
B
${10^5}\,\,K$
C
${10^3}\,\,K$
D
${10^9}\,\,K$

Explanation

The average kinetic energy per molecule $= {3 \over 2}kT$

This kinetic energy should be able to provide the repulsive potential energy

$\therefore$ ${3 \over 2}kT = 7.7 \times {10^{ - 14}}$

$\Rightarrow T = {{2 \times 7.7 \times {{10}^{ - 14}}} \over {3 \times 1.38 \times {{10}^{ - 23}}}} = 3.7 \times {10^9}$
2

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
3

AIEEE 2003

Which of the following atoms has the lowest ionization potential ?
A
${}_7^{14}N$
B
${}_{55}^{133}\,Cs$
C
${}_{18}^{40}\,Ar$
D
${}_8^{16}\,O$

Explanation

The ionisation potential increases from left to right in a period and decreases from top to bottom in a group. Therefore ceasium will have the lowest ionisation potential.
4

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

$92-16+4-2=78$