### JEE Mains Previous Years Questions with Solutions

4.5
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1

### AIEEE 2004

An $\alpha$-particle of energy $5$ $MeV$ is scattered through ${180^ \circ }$ by a fixed uranium nucleus. The distance of closest approach is of the order of
A
${10^{ - 12}}\,cm$
B
${10^{ - 10}}\,cm$
C
$1A$
D
${10^{ - 15a}}\,cm$

## Explanation

KEY NOTE :

Distance of closest approach

${r_0} = {{Ze\left( {2e} \right)} \over {4\pi {\varepsilon _0}E}}$

Energy, $E = 5 \times {10^6} \times1.6 \times {10^{ - 19}}J$

$\therefore$ ${r_0} = {{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)} \over {5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}$

$\Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm$
2

### AIEEE 2003

If the binding energy of the electron in a hydrogen atom is $13.6eV,$ the energy required to remove the electron from the first excited state of $L{i^{ + + }}$ is
A
$30.6$ $eV$
B
$13.6$ $eV$
C
$3.4$ $eV$
D
$122.4$ $eV$

## Explanation

${E_n} = - {{13.6} \over {{n^2}}}{Z^2}eV/$atom

For lithium ion $Z=3;$ $n=2$ (for first excited state)

${E_n} = - {{13.6} \over {{2^2}}} \times {3^2} = - 30.6eV$
3

### AIEEE 2003

Which of the following cannot be emitted by radioactive substances during their decay ?
A
Protons
B
Neutrinoes
C
Helium nuclei
D
Electrons

## Explanation

The radioactive substances emit $\alpha$ -particles (Helium nucleus), $\beta$ -particles (electrons) and neutrinoes.
4

### 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

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

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