1

### 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
2

### 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.
3

### 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
4

### AIPMT 2004

The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in hydrogen atom will be (Given ionization energy of H = 2.18 $\times$ 10$-$18 J atom$-$1 and h = 6.625 $\times$ 10$-$34 J s)
A
1.54 $\times$ 1015 s$-$1
B
1.03 $\times$ 1015 s$-$1
C
3.08 $\times$ 1015 s$-$1
D
2.00 $\times$ 1015 s$-$1

## Explanation

E = h$v$ or $v$ = E/h

For H atom, E = ${{ - 21.76 \times {{10}^{ - 19}}} \over {{n^2}}}J\,at{m^{ - 1}}$

$\Delta E = - 21.76 \times {10^{ - 19}}\left( {{1 \over {{4^2}}} - {1 \over {{1^2}}}} \right)$

= 20.40 $\times$ 10-19 J atm-1

$v$ = ${{20.40 \times {{10}^{ - 19}}} \over {6.626 \times {{10}^{ - 34}}}}$ = 3.079 $\times$ 1015 s-1