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### 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 $$\,\,\, \lambda = {{36A} \over 7} 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with : A Lyman series B Balmer series C Paschen series D Brackett series ## Explanation Here electron is coming to the orbital of radius 211.6 pm. Now we have to find which series have radius of orbital 211.6 pm. We know, Radius, r = 0.529 \times {{{n^2}} \over Z}\,\mathop A\limits^ \circ Given, r = 211.6 pm = 211.6 \times 10-12 m and Z = 1 for hydrogen atom \therefore 211.6 \times 10-12 = 0.529 \times {{{n^2}} \over 1} \times 10-10 \Rightarrow n = 2 As n = 2 so the series is Balmer Series. 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Morning Slot Ejection of the photoelectron from metal in the photoelectric experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is : A 4 eV B 4.5 eV C 5 eV D 5.5 eV ## Explanation \lambda = 250 nm = 2500 \mathop A\limits^ \circ E = {{hc} \over \lambda } = {{12400} \over {2500}} = 4.96 eV K. E = 0.5 eV As, \,\,\,\,\,\, K. E = E - \omega 0 \Rightarrow \,\,\, 0.5 = 4.96 - {\omega _0} \Rightarrow \,\,\, {\omega _0} = 4.46 eV \simeq 4.5 eV 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Evening Slot The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is : A 0.529 \mathop {\rm A}\limits^ \circ B 2\pi \times 0.529 \mathop {\rm A}\limits^ \circ C {{0.529} \over {2\pi }} \mathop {\rm A}\limits^ \circ D 4 \times 0.529 \mathop {\rm A}\limits^ \circ ## Explanation Radius r = 0.529 \times {{{n^2}} \over z} In first Bohr orbit of hydrogen atom radius, r = 0.529 \times {{{1^2}} \over 1} = 0.529 \mathop A\limits^ \circ Angular momentum, mvr = {{nh} \over {2\pi }} \therefore\,\,\, for n = 1, mvr = {h \over {2\pi }} \Rightarrow$$\,\,\,$ 2$\pi$r = ${h \over {mv}}$ = $\lambda$   [ as   $\lambda$ = ${h \over {mv}}$]

$\therefore\,\,\,$ $\lambda$ = 2$\pi$r = 2 $\pi$ $\times$ 0.5 29