JEE Mains Previous Years Questions with Solutions

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1

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 $\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}$
$= 1.214 \times {10^{ - 7}}m$
2

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.
3

AIEEE 2011

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
4

AIEEE 2010

Ionisation energy of He+ is 19.6 x 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is
A
4.41 x 10–16 J atom–1
B
-4.41 x 10–17 J atom–1
C
-2.2 x 10–15 J atom–1
D
8.82 x 10–17 J atom–1

Explanation

Ionisation energy(IE) - It is the energy required to move an electron from ground state to infinity.

IE = ${E_\infty } - {E_1}$ = $0 - {E_1}$ = $- {E_1}$

$\therefore$ E1 of He+ = - 19.6 x 10–18 J atom–1

Energy of a species at n state,

(En)species = (En)hydrogen $\times$ Z2

$\therefore$ (E1)hydrogen = ${{ - 19.6 \times {{10}^{ - 18}}} \over 4}$ [ For He, Z = 2 ]

(E1)Li+2 = ${{ - 19.6 \times {{10}^{ - 18}}} \over 4} \times {3^2}$

= -4.41 x 10–17 J atom–1