1

### 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
2

### 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
3

### AIPMT 2003

The velocity of Planck's constant is 6.63 $\times$ 10$-$34 J s.

The velocity of light is 3.0 $\times$ 108 m s$-$1. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8 $\times$ 1015 s$-$1 ?
A
2 $\times$ 10$-$25
B
5 $\times$ 10$-$18
C
$4 \times {10^1}$
D
3 $\times$ 107

## Explanation

$v$ = c/$\lambda$

$\lambda$ = ${c \over v}$ = ${{3 \times {{10}^8}} \over {8 \times {{10}^{15}}}}$ = 37.5 $\times$ 10-9 m

= 37.5 nm $\approx$ $4 \times {10^1}$ nm
4

### AIPMT 2002

In hydrogen atom, energy of first excited state is $-$ 3.4 eV. Then find out K.E. of same orbit of hydrogen atom
A
+3.4 eV
B
+6.8 eV
C
$-$ 13.6 eV
D
+13.6 eV

## Explanation

K.E = 1/2 mv2

= ${\left( {{{\pi {e^2}} \over {nh}}} \right)^2} \times 2m$ [ $\because$ v = ${{2\pi {e^2}} \over {nh}}$]

$\therefore$ Total energy = En = ${{ - 2{\pi ^2}m{e^4}} \over {{n^2}{h^2}}}$

= -${\left( {{{\pi {e^2}} \over {nh}}} \right)^2}$ $\times$ 2m = -K.E

$\therefore$ K.E = - En

Energy of first excited state is -3.4 eV

$\therefore$ Kinetic energy of the same orbit (n = 2) will be +3.4 ev