1

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
2

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
3

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.
4

AIPMT 2011 Prelims

If n = 6, the correct sequence for filling of electrons will be
A
$ns \to \left( {n - 2} \right)f \to \left( {n - 1} \right)d \to np$
B
$ns \to \left( {n - 1} \right)d \to \left( {n - 2} \right)f \to np$
C
$ns \to \left( {n - 2} \right)f \to np \to \left( {n - 1} \right)d$
D
$ns \to np\left( {n - 1} \right)d \to \left( {n - 2} \right)f$