1
MCQ (Single Correct Answer)

### AIPMT 2007

Consider the following sets of quantum numbers :

n l m s
(i) 3 0 0 +1/2
(ii) 2 2 1 +1/2
(iii) 4 3 -2 -1/2
(iv) 1 0 -1 -1/2
(v) 3 2 3 +1/2

Which of the following sets of quantum number is not possible ?
A
(i), (ii), (iii) and (iv)
B
(ii), (iv) and (v)
C
(i) and (iii)
D
(ii), (iii) and (iv)

## Explanation

(i) represents an electron in 3s orbital

(ii) is not possible as value of m varies from 0, 1, .... ($n$ -1)

(iii) represents an electron in 4f orbital

(iv) is not possible as value of m varies from -$l$ ... +$l$

(v) is not possible as value of m varies from -$l$ ... +$l$, it can never be grater than $l$
2
MCQ (Single Correct Answer)

### AIPMT 2006

Given : The mass of electron is 9.11 $\times$ 10$-$31 kg, Planck constant is 6.626 $\times$ 10$-$34 J s, the uncertainty involved in the measurement of velocity within a distance of 0.1 $\mathop A\limits^ \circ$ is
A
5.79 $\times$ 105 m s$-$1
B
5.79 $\times$ 106 m s$-$1
C
5.79 $\times$ 107 m s$-$1
D
5.79 $\times$ 108 m s$-$1

## Explanation

$\Delta x.m\Delta v = h/4\pi$

0.1 $\times$ 10-10 $\times$ 9.11 $\times$ 10-31 $\times$ $\Delta v$ = ${{6.626 \times {{10}^{ - 34}}} \over {4 \times 3.143}}$

$\therefore$ $\Delta v$ = ${{6.626 \times {{10}^{ - 34}}} \over {4 \times 3.143 \times 0.1 \times {{10}^{ - 10}} \times 9.11 \times {{10}^{ - 31}}}}$

= 5.79 $\times$ 106 m s-1
3
MCQ (Single Correct Answer)

### AIPMT 2006

The orientation of an atomic orbital is governed by
A
principal quantum number
B
azimuthal quantum number
C
spin quantum number
D
magnetic quantum number.

## Explanation

Magnetic quantum number describes the orientation.
4
MCQ (Single Correct Answer)

### AIPMT 2005

The energy of second Bohr orbit of the hydrogen atom is $-$328 kJ mol$-$1; hence the energy of fourth Bohr orbit would be
A
$-$ 41 kJ mol$-$1
B
$-$82 kJ mol$-$1
C
$-$164 kJ mol$-$1
D
$-$1312 kJ mol$-$1

## Explanation

En = -K${\left( {{Z \over n}} \right)^2}$

Z = 1; n = 2

E2 = ${ {{-K \times 1 \over 4}}}$$\Rightarrow$ E2 = -328 kJ mol-1;

K = 4 $\times$ 328

E4 = ${ {{-K \times 1 \over 16}}}$ $\Rightarrow$ E4 = -4 $\times$ 328 ${ {{ 1 \over 16}}}$

= -82 kJ mol-1

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