### JEE Mains Previous Years Questions with Solutions

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1

### 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
2

### AIEEE 2009

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
3

### AIEEE 2008

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
4

### AIEEE 2008

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.