1

### 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
2

### 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 = [Xe]4f7 5d1 6s2
3

### 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
4

### NEET 2013

Based on equation E = $-$ 2.178 $\times$ 10$-$18 J $\left( {{{{Z^2}} \over {{n^2}}}} \right)$, certain conclusions are written. Which of them is not correct ?
A
Equation can be used to calculate the change in energy when the electron changes orbit.
B
For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.
C
The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.
D
Larger the value of n, the larger is the orbit radius.

## Explanation

The electron is more tightly bound in the smallest allowed orbit.