1

### 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
2

### 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.
3

### NEET 2013

What is the maximum numbers of electrons that can be associated with the following set of quantum numbers ?
n = 3, $l$ = 1 and m = $-$ 1
A
4
B
2
C
10
D
6

## Explanation

The orbitals associated with n = 3, l = 1 is 3p. One orbital (with m = -1) of 3p-subshell can only accomodate maximum 2 electrons.
4

### 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 = $\sqrt {l(l + 1)} {h \over {2\pi }}$ = ${ \sqrt 2 {h}} \over 2 \pi$ = ${h \over {\sqrt 2 \pi }}$