 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2006

If the binding energy per nucleon in ${}_3^7Li$ and ${}_2^4He$ nuclei are $5.60$ $MeV$ and $7.06$ $MeV$ respectively, then in the reaction $$p + {}_3^7Li \to 2\,{}_2^4He$$
energy of proton must be
A
$28.24$ $MeV$
B
$17.28$ $MeV$
C
$1.46$ $MeV$
D
$39.2$ $MeV$

Explanation

Let $E$ be the energy of proton, then

$E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right]$

$\Rightarrow E = 56.48 - 39.2 = 17.28MeV$
2

AIEEE 2006

When ${}_3L{i^7}$ nuclei are bombarded by protons, and the resultant nuclei are ${}_4B{e^8}$, the emitted particles will be
A
alpha particles
B
beta particles
C
gamma photons
D
neutrons

Explanation

${}_3^7Li + {}_1^1p \to {}_4^8Be + {}_0^0\gamma$
3

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 $r \propto {1 \over m}.$
4

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.