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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
The correct set of four quantum numbers for the valence elections of rubidium atom (Z= 37) is:
A
5, 1, 1, + 1/2
B
5, 1, 0, + 1/2
C
5, 0, 0, + 1/2
D
5, 0, 1, + 1/2

Explanation

The electronic configuration of Rubidium (Rb = 37):
1s22s22p63s23p63d104s24p65s1

As you can see last electron or valence electron enterin 5s subshell

So, the quantum numbers are n = 5, $$l$$ = 0, m = 0, s = $$ \pm {1 \over 2}$$
2

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
Energy of an electron is given by $$E = - 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)$$. Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be
(h = 6.62 × 10−34 Js and c = 3.0 × 108 ms−1)
A
2.816 × 10−7 m
B
6.500 × 10−7 m
C
8.500 × 10−7 m
D
1.214 × 10−7 m

Explanation

Given $$E$$ = $$- 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)$$
                = $$ - 2.178 \times {10^{ - 8}}{Z^2}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$$
Electron in an hydrogen atom exited from level n = 1 to n = 2. So n2 = 2 and n1 = 1.
And for H, Z = 1
$$\therefore$$ $$E = - 2.178 \times {10^{ - 8}}\left( {{1 \over {{{\left( 2 \right)}^2}}} - {1 \over {{{\left( 1 \right)}^2}}}} \right)$$
           = 1.6335 $$ \times $$ 10-18 J

We know $$E$$ = $${{hc} \over \lambda }$$
$$\therefore$$ $${{hc} \over \lambda }$$ = 1.6335 $$ \times $$ 10-18
then $$\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}$$
             $$ = 1.214 \times {10^{ - 7}}m$$
3

AIEEE 2012

MCQ (Single Correct Answer)
The electrons identified by quantum numbers n and l :
(a) n = 4, $$l$$ = 1
(b) n = 4, $$ l$$ = 0
(c) n = 3, $$l$$ = 2
(d) n = 3, $$l$$ = 1
Can be placed in order of increasing energy as :
A
(c) < (d) < (b) < (a)
B
(d) < (b) < (c) < (a)
C
(b) < (d) < (a) < (c)
D
(a) < (c) < (b) < (d)

Explanation

(a) n = 4, $$l$$ = 1 (p-subshell), so 4p
(b) n = 4, $$l$$ = 0 (s-subshell), so 4s
(c) n = 3, $$l$$ = 2 (d-subshell), so 3d
(d) n = 3, $$l$$ = 1 (p-subshell), so 3p

Accroding to the Bohr ( n + $$l$$ ) rule,
Enery order of the subshell : 3p < 4s < 3d < 4p

Note: When two orbital have the same value of ( n + $$l$$ ) then the orbital which have lower value of n will be filled first.
4

AIEEE 2011

MCQ (Single Correct Answer)
A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :
A
325 nm
B
743 nm
C
518 nm
D
1035 nm

Explanation

Here a photon is divided into two photons of different wavelength.

According to the law of conservation of energy,

Energy of photon before dividing into two parts = Energy of first photon + Energy of second photon

$${{hc} \over \lambda } = {{hc} \over {{\lambda _1}}} + {{hc} \over {{\lambda _2}}}$$

$$ \Rightarrow {1 \over \lambda } = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$

$$ \Rightarrow {1 \over {355}} = {1 \over {680}} + {1 \over {{\lambda _2}}}$$

$${\lambda _2} = 742.77$$ nm = 743 nm

Questions Asked from Structure of Atom

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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