1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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.

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