1

### AIPMT 2012 Mains

The orbital angular momentum of a p-electron is given as
A
${h \over {\sqrt 2 \pi }}$
B
$\sqrt 3 {h \over {2\pi }}$
C
$\sqrt {{3 \over 2}} {h \over \pi }$
D
$\sqrt 6 {h \over {2\pi }}$

## Explanation

Orbital angular momentum (m) = $\sqrt {l(l + 1)} {h \over {2\pi }}$

For p-electrons; l = 1

Thus, m = $\sqrt {l(l + 1)} {h \over {2\pi }}$ = ${ \sqrt 2 {h}} \over 2 \pi$ = ${h \over {\sqrt 2 \pi }}$
2

### AIPMT 2012 Prelims

The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is
A
5, 1, 1, + 1/2
B
6, 0, 0, +1/2
C
5, 0, 0, +1/2
D
5, 1, 0, +1/2

## Explanation

Rb(37) = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1

For 5s, n= 5, l = 0, m = 0, s = +1/2 or -1/2
3

### AIPMT 2012 Prelims

Maximum number of electrons in a subshell with $l$ = 3 and n = 4 is
A
14
B
16
C
10
D
12

## Explanation

l = 3 and n = 4 represent 4f, So total number of electrons = 2(2$l$+1) = 14
4

### AIPMT 2011 Mains

According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon ?
A
n = 6 to n = 1
B
n = 5 to n = 4
C
n = 6 to n = 5
D
n = 5 to n = 3

## Explanation

We know that

$\Delta E \propto \left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$, where n2 > n1

$\therefore$ n = 6 and n = 5 will give least energetic photon.