JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

An alpha nucleus of energy ${1 \over 2}m{v^2}$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to
A
${v^2}$
B
${1 \over m}$
C
${1 \over {{v^2}}}$
D
${1 \over {Ze}}$

Explanation

Work done to stop the $\alpha$ particle is equal to $K.E.$

$\therefore$ $qV = {1 \over 2}m{v^2} \Rightarrow q \times {{K\left( {Ze} \right)} \over r} = {1 \over 2}m{v^2}$

$\Rightarrow r = {{2\left( {2e} \right)K\left( {Ze} \right)} \over {m{V^2}}} = {{4KZ{e^2}} \over {m{v^2}}}$

$\Rightarrow r \propto {1 \over {{v^2}}}$ and $r \propto {1 \over m}.$
2

AIEEE 2005

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
A
$iv$
B
$iii$
C
$ii$
D
$i$

Explanation

KEY CONCEPT : $E = Rhc\left[ {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right]$

$E$ will be maximum for the transition for which

$\left[ {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right]$ is maximum. Here ${n_2}$ is the higher energy level

Clearly, $\left[ {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right]$ is maximum for the third

transition, i.e. $2 \to 1.$ $I$ transition represents the absorption of energy.
3

AIEEE 2005

A nuclear transformation is denoted by $X\left( {n,\alpha } \right)\matrix{ 7 \cr 3 \cr } Li.$ Which of the following is the nucleus of element $X$ ?
A
${}_5^{10}B$
B
${}^{12}{C_6}$
C
${}_4^{11}Be$
D
${}_5^9B$

Explanation

${}_z{X^A} + {}_0{n^1} \to {}_3L{i^7} + {}_2H{e^4}$

On comparison,

$A = 7 + 4 - 1 = 10,\,\,\,z = 3 + 2 - 0 = 5$

It is boron ${}_5{B^{10}}$
4

AIEEE 2005

The intensity of gamma radiation from a given source is $L$. On passing through $36$ $mm$ of lead, it is reduced to ${{\rm I} \over 8}.$ The thickness of lead which will reduce the intensity to ${{\rm I} \over 2}$ will be
A
$9mm$
B
$6mm$
C
$12mm$
D
$18mm$

Explanation

KEY CONCEPT : Intensity $I = {I_0}.{e^{ - \mu d}},$

Applying logarithm on both sides,

$- \mu d = \log \left( {{I \over {{I_0}}}} \right)$

$- \mu \times 36 = \log \left( {{{I/8} \over I}} \right).........\left( i \right)$

$- \mu \times d = \log \left( {{{I/2} \over I}} \right).........\left( {ii} \right)$

Dividing $(i)$ by $(ii),$

${{36} \over d} = {{\log \left( {{1 \over 8}} \right)} \over {\log \left( {{1 \over 2}} \right)}}$

$= {{3\log \left( {{1 \over 2}} \right)} \over {\log \left( {{1 \over 2}} \right)}} = 3$

or $\,\,\,d = {{36} \over 3} = 12\,mm$