1

### AIPMT 2014

Calculate the energy in joule corresponding to light of wavelength 45 nm. (Planck,s constant, h = 6.63 $\times$ 10$-$34 J s, speed of light, c = 3 $\times$ 108 m s$-$1)
A
6.67 $\times$ 1015
B
6.67 $\times$ 1011
C
4.42 $\times$ 10$-$15
D
4.42 $\times$ 10$-$18

## Explanation

E = $hc \over \lambda$

E = ${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {45 \times {{10}^{ - 9}}}}$

= 4.42 $\times$ 10-18 J
2

### 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
3

### 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
4

### 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