 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

JEE Main 2015 (Offline)

Which of the following is the energy of a possible excited state of hydrogen?
A
–6.8 eV
B
–3.4 eV
C
+6.8 eV
D
+13.6 eV

Explanation

Energy(En) = $- {{13.6{Z^2}} \over {{n^2}}}eV$

where n = 1, 2, 3, 4, ......

For hydrogen Z = 1 and excited state starts from n = 2

So putting n = 2

E2 = $- {{13.6} \over {{2^2}}}eV$ = -3.4 eV
2

JEE Main 2014 (Offline)

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 = $\pm {1 \over 2}$
3

JEE Main 2013 (Offline)

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 $\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}$
$= 1.214 \times {10^{ - 7}}m$
4

AIEEE 2012

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.