1

AIPMT 2014

What is the maximum number of orbitals that can be identified with the following quantum numbers ?
n = 3, l = 1, m1 = 0
A
1
B
2
C
3
D
4

Explanation

Only one orbital 3p has following set of quantum numbers, n = 3, l = 1 and ml = 0
2

NEET 2013 (Karnataka)

According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as (h = 6.62 $\times$ 10$-$27 ergs, c = 3 $\times$ 1010 cm s$-$1, NA = 6.02 $\times$ 10$-$23 mol$-$1)
A
${{1.196 \times {{10}^8}} \over \lambda }$
B
${{2.859 \times {{10}^5}} \over \lambda }$
C
${{2.859 \times {{10}^{16}}} \over \lambda }$
D
${{1.196 \times {{10}^{16}}} \over \lambda }$

Explanation

E = ${{hc{N_A}} \over \lambda }$

= ${{6.62 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}} \times 6.02 \times {{10}^{23}}} \over \lambda }$

= ${{1.1955 \times {{10}^8}} \over \lambda }$ = ${{1.196 \times {{10}^8}} \over \lambda }$ ergs mol-1
3

NEET 2013 (Karnataka)

The outer electronic configuration of Gd (At. No. 64) is
A
4f5 5d4 6s1
B
4f7 5d1 6s2
C
4f3 5d5 6s2
D
4f4 5d5 6s1

Explanation

Gd[64] = [Xe]4f7 5d1 6s2
4

NEET 2013

The value of Planck's constant is 6.63 $\times$ 10$-$34 J s. The speed of light is 3 $\times$ 1017 mm s$-$1. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6 $\times$ 1015 s$-$1 ?
A
50
B
75
C
10
D
25

Explanation

c = $\nu \lambda$

$\lambda$ = ${c \over \nu }$

= ${{3 \times {{10}^{17}}} \over {6 \times {{10}^{15}}}}$ = 50 nm