### JEE Mains Previous Years Questions with Solutions

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1

### 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
2

### 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
3

### 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
4

### AIEEE 2009

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 $\times$ 10-34 kg m2s-1, mass of electron, em = 9.1 $\times$ 10-31 kg)
A
5.10 $\times$ 10-3 m
B
1.92 $\times$ 10-3 m
C
3.84 $\times$ 10-3 m
D
1.52 $\times$ 10-4 m

## Explanation

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

$\therefore$ 0.005 = ${{\Delta V} \over {600}} \times 100$

$\Rightarrow$ $\Delta$V = 3 $\times$ 10-2

According to Heisenberg uncertainty principle,

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

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

$\Rightarrow$ $\Delta x = {{6.63 \times {{10}^{ - 34}}} \over {4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^{ - 2}}}}$

= 1.92 $\times$ 10-3 m