### JEE Mains Previous Years Questions with Solutions

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1

### 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.
2

### 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
3

### 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
4

### AIEEE 2010

The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl – Cl bond is
(c = 3 x 108 ms–1 and NA = 6.02 x 1023 mol–1)
A
594 nm
B
640 nm
C
700 nm
D
494 nm

## Explanation

Energy required to break one Cl2 molecule = ${{242 \times {{10}^3}} \over {6.02 \times {{10}^{23}}}}$ J

As E = ${{hc} \over \lambda }$

So $\lambda = {{hc} \over E}$

= ${{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 6.02 \times {{10}^{23}}} \over {242 \times {{10}^3}}}$

= 494 $\times$ 10-9 m

= 494 nm