1
MCQ (Single Correct Answer)

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

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