### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2006

Uncertainty in the position of an electron (mass = 9.1 $\times$ 10-31 kg) moving with a velocity 300 ms-1, accurate upto 0.001% will be (h = 6.63 $\times$ 10-34 Js)
A
1.92 $\times$ 10-2 m
B
3.84 $\times$ 10-2 m
C
19.2 $\times$ 10-2 m
D
5.76 $\times$ 10-2 m

## Explanation

% error in velocity = ${{\Delta V} \over V} \times 100$

$\therefore$ 0.001 = ${{\Delta V} \over {300}} \times 100$

$\Rightarrow$ $\Delta$V = 3 $\times$ 10-3

According to Heisenberg uncertainty principle,

$\Delta x.m\Delta V \ge {h \over {4\pi }}$

$\Rightarrow$ $\Delta x = {h \over {4\pi m\Delta V}}$

$\Rightarrow$ $\Delta x = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^{ - 3}}}}$

= 1.92 $\times$ 10-2 m
2

### AIEEE 2006

According to Bohr's theory, the angular momentum of an electron in 5th orbit is
A
10 $h/\pi$
B
2.5 $h/\pi$
C
25 $h/\pi$
D
1.0 $h/\pi$

## Explanation

Formula of angular momentum of an electron,

mvr = ${{nh} \over {2\pi }}$

here n = 5

$\therefore$ mvr = ${{5h} \over {2\pi }}$ = 2.5${{h} \over {\pi }}$
3

### AIEEE 2005

Pick out the isoelectronic structure from the following :

\eqalign{ & \left( i \right)\,\,\,\,\,\,C{H_3}^ + \cr & \left( {ii} \right)\,\,\,\,{H_3}{O^ + } \cr & \left( {iii} \right)\,\,\,N{H_3} \cr & \left( {iv} \right)\,\,\,\,C{H_3}^ - \cr}
A
$(i)$ and $(ii)$
B
$(iii)$ and $(iv)$
C
$(i)$ and $(iii)$
D
$(ii), (iii)$ and $(iv)$

## Explanation

Ions No. of electrons
CH3+ 8
H3O+ 10
NH3 10
CH3- 10

$\therefore$ $\,\,\,$ ${H_3}{O^ + },N{H_3}$ and $C{H_3}^ -$ are isoelectronic.
4

### AIEEE 2005

In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields?
(A) n = 1, l = 0, m = 0
(B) n = 2, l = 0, m = 0
(C) n = 2, l = 1, m = 1
(D) n = 3, l = 2, m = 1
(E) n = 3, l = 2, m = 0
A
(D) and (E)
B
(C) and (D)
C
(B) and (C)
D
(A) and (B)

## Explanation

As here is no electric and magnetic field so ignore m to calculate the energy of orbital.

As here atom is multi-electron so ( n + l ) rule is applicable here. This rule says those orbitals which have more value of ( n + l ) will have more energy.

In (A), n + l = 1 + 0 = 1

In (B), n + l = 2 + 0 = 2

In (C), n + l = 2 + 1 = 3

In (D), n + l = 3 + 2 = 5

In (E), n + l = 3 + 2 = 5

So (D) and (E) will be of equal energy.