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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2009

MCQ (Single Correct Answer)
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 x 103 ms−1 (Mass of proton = 1.67 $$\times$$ 10-27 kg and h = 6.63 $$\times$$ 10-34 Js) :
A
0.40 nm
B
2.5 nm
C
14.0 nm
D
0.32 nm

Explanation

Wavelength$$\left( \lambda \right)$$ = $${h \over {mv}}$$

= $${{6.63 \times {{10}^{ - 34}}} \over {1.67 \times {{10}^{ - 27}} \times {{10}^3}}}$$

= 0.4 $$\times$$ 10-9

= 0.4 nm
2

AIEEE 2008

MCQ (Single Correct Answer)
The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol−1. The energy required to excite the electron in the atom from n = 1 to n = 2 is
A
8.51 × 105 J mol−1
B
6.56 × 105 J mol−1
C
7.56 × 105 J mol−1
D
9.84 × 105 J mol−1

Explanation

Note :

1 eV/atom = 96.485 $$ \times $$ 103 J/mol

$$\therefore$$ 13.6 eV/atom = 13.6$$ \times $$ 96.485 $$ \times $$ 103 J/mol = 1.312 × 106 J mol−1

Energy required to excite the electron from n1 to n2 is

$$\Delta E = 13.6 \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$

= 1.312 × 106 × 1$$\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$$

= 1.312 × $${3 \over 4}$$ × 106

= 9.84 × 105 J mol−1
3

AIEEE 2008

MCQ (Single Correct Answer)
Which one of the following constitutes a group of the isoelectronic species?
A
$$C_2^{2 - }$$, $$O_2^{-}$$, CO, NO
B
$$NO^{+}$$, $$C_2^{2 - }$$, CN-, $$N_2$$
C
CN-, $$N_2$$, $$O_2^{2-}$$, $$C_2^{2 - }$$
D
$$N_2$$, $$O_2^{-}$$, $$NO^{+}$$, CO

Explanation

Species No. of electrons
C22- 12+2 = 14
O2- 16+1 = 17
CO 6+8 = 14
NO 7+8 = 15
NO+ 7+8-1 = 14
C22- 12+2 = 14
CN- 6+7+1 = 14
N2 14
CN- 6+7+1 = 14
N2 14
O22- 16+2 = 18
C2 12
N2 14
O2- 16+1 = 17
NO+ 7+8-1 = 14
CO 6+8 = 14


So, option $$(B)$$ is correct.
4

AIEEE 2007

MCQ (Single Correct Answer)
Which of the following sets of quantum numbers represents the highest energy of an atom?
A
n = 3, l = 0, m = 0, s = +1/ 2
B
n = 3, l = 1, m = 1, s = +1/ 2
C
n = 3, l = 2, m = 1, s = +1/ 2
D
n = 4, l = 0, m = 0, s = +1/ 2

Explanation

In which quantum number have highest ( n + l ) value, will represent highest energy of an atom.

In option (C), n + l = 3 + 2 = 5 = maximum.

Questions Asked from Structure of Atom

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