1

### JEE Main 2016 (Online) 9th April Morning Slot

The total number of orbitals associated with the principal quantum number 5 is :
A
5
B
10
C
20
D
25
2

### JEE Main 2017 (Offline)

The radius of the second Bohr orbit for hydrogen atom is:
(Planck’s Const. h = 6.6262 × 10-34 Js; mass of electron = 9.1091 × 10-31 kg; charge of electron (e) = 1.60210 × 10-19 C; permittivity of vacuum (${\varepsilon _0}$) = 8.854185 × 10-12 kg-1 m-3 A2)
A
4.76 $\mathop {\rm A}\limits^o$
B
2.12 $\mathop {\rm A}\limits^o$
C
0.529 $\mathop {\rm A}\limits^o$
D
1.65 $\mathop {\rm A}\limits^o$

## Explanation

Radius of an atom in nth orbit,

rn = 0.529 $\times {{{n^2}} \over Z}$

Here n = 2

and for hydrogen atom, atomic number (Z) = 1

$\therefore$ r2 = 0.529$\times {{{2^2}} \over 1}$ = 2.12 $\mathop {\rm A}\limits^o$
3

### JEE Main 2017 (Offline)

The group having isoelectronic species is:
A
O2- , F- , Na+, Mg2+
B
O2- , F- , Na, Mg2+
C
O- , F- , Na+, Mg2+
D
O- , F- , Na, Mg+

## Explanation

O2- , F- , Na+ and Mg2+, all have 10 electrons each.
4

### JEE Main 2017 (Online) 8th April Morning Slot

If the shortest wavelength in Lyman series of hydrogen atom is A, then the longestwavelength in Paschen series of He+ is :
A
${{5A} \over 9}$
B
${{9A} \over 5}$
C
${{36A} \over 5}$
D
${{36A} \over 7}$

## Explanation

Note :

(1)   In Lyman Series, transition happens in n = 1 state
from n = 2, 3, . . . . . $\propto$

(2)   In Balmer Series, transition happens in n = 2 state
from n = 3, 4, . . . . . $\propto$

(3)   In Paschen Series, transition happens in n = 3 state
from n = 4, 5, . . . . . $\propto$

(4)   In Bracktt Series, transition happens in n = 4 state
from n = 5, 6 . . . . . . $\propto$

(5)   In Pfund Series, transition happens in n = 5 state
from n = 6, 7, . . . . $\propto$

We know,

${1 \over \lambda }$ = Rz2 (${1 \over {n_1^2}}$ $-$ ${1 \over {n_2^2}}$)

The shortest wavelength of hydrogen atom in Lyman series is from n1 = 1 to n2 = $\propto$

$\therefore\,\,\,$ ${1 \over A}$ = 12 R (${1 \over {{1^2}}}$ $-$ ${1 \over {{ \propto ^2}}}$)    [for hydrogen, z = 1]

$\Rightarrow$ $\,\,\,$ ${1 \over A}$ = R

The longest wavelength in pascal series of He+ is from n1 = 3 to n2 = 4

For He+, z = 2.

${1 \over \lambda }$ = RZ2 (${1 \over {n_1^2}}$ $-$ ${1 \over {n_2^2}}$)

$\Rightarrow $$\,\,\, {1 \over \lambda } = {1 \over A} (2)2 ({1 \over {{3^2}}} - {1 \over {{4^2}}}) \Rightarrow$$\,\,\,$ ${1 \over \lambda }$ = ${4 \over A}$ $\left( {{7 \over {16 \times 9}}} \right)$ = ${7 \over {36A}}$

$\Rightarrow$$\,\,\,$ $\lambda$ = ${{36A} \over 7}$