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The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with :
Option: 1 Lyman series
Option: 2 Balmer series
Option: 3 Paschen series
Option: 4 Brackett series

For hydrogen atom, the radius of n^??????th orbit, r_??????n= 52.9 pm ×n²

211.6 pm = 52.9 pm×n²

n^??????2=4

n=2

Hence, the transition is from a higher orbit to the second orbit.
This corresponds to the Balmer series.

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vishal kumar

For the Balmer series in the spectrum of H atom, $\bar{v}=\mathrm{R}_{\mathrm{H}}\left\{\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right\}$ the correct statements among (I) to (IV) are: (1) As wavelength decreases, the lines in the series converge
(II) The integer n₁ is equal to 2
(III) The lines of longest wavelength corresponds to n₂ = 3
(IV) The ionization energy of hydrogen can be calculated from wavenumber of these lines

Option: 1 $(II),(III),(IV)$
Option: 2 $(I),(III),(IV)$
Option: 3 $(I),(II),(III)$
Option: 4 $(I),(II),(IV)$

Line Spectrum of Hydrogen-like atoms

$\frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )$

Where R is called Rydberg constant, R = 1.097 X 10⁷, Z is the atomic number

n₁= 1, 2, 3….

n₂= n₁+1, n₁+2 ……

Lyman Series spectrum:

Where

n₁= 1 and n₂= 2, 3, 4....

This lies in the Ultraviolet region.

Balmer Series Spectrum:

Where n₁= 2 and n₂= 3, 4, 5, 6....

It lies in the visible region.

Paschen, Bracket and Pfund Series spectrums:

these lies in Infrared Region.

Since, $\mathrm{\Delta E\: \propto\:\frac{1}{\lambda} }$ so for wavelength to be longest or maximum the energy gap should be minimum. For the Balmer series, the transition for the longest wavelength is from n = 3 to n = 2. Thus clearly, n₁ value for the Balmer series is 2. Further, as wavelength decreases, the lines in the series converge.

Therefore, Option(3) is correct.

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vishal kumar

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The number of orbitals associated with quantum numbers $n = 5, m_s = +1/2$
is:
Option: 1 25
Option: 2 11
Option: 3 15
Option: 4 50

We have:

n = 5, m_s = +1/2

Thus, the values of l are from 0 to (n-1)

l = 0 to 4

Thus, values of l are 5s, 5p, 5d, 5f, and 5g

Now, the total number of orbitals $= n^2 = 5^2 = 25$

Therefore, Option(1) is correct.

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Kuldeep Maurya

The radius of the second Bohr orbit, in terms of the Bohr radius, $a_{0},$ in $Li^{2+}$ is :
Option: 1 $\frac{2a_{0}}{3}$

Option: 2 $\frac{4a_{0}}{3}$

Option: 3 $\frac{4a_{0}}{9}$

Option: 4 $\frac{2a_{0}}{9}$

As we have learnt,

Radius, velocity and the energy of nth Bohr orbital -

Bohr radius of n^th orbit:

$r_{n}= 0.529 \frac{n^{2}}{z}A^{0}$

where Z is the atomic number

Velocity of electron in n^thorbit:

$v_{n}= (2.165\times 10^{6})\frac{z}{n}\: m/s$

where z is atomic number

Total energy of electron in n^th orbit:

$E_{n}= -13.6\: \frac{z^{2}}{n^{2}}eV$

Where z is atomic number

$r=\frac{n^2a_o}{Z}$ .

$\mathrm{For\ Li^{2+}\ r=\frac{2^2a_o}{3}=\frac{4a_o}{3}}$

Therefore, Option(2) is correct.

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vishal kumar

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In the sixth period, the orbitals that are filled are :
Option: 1 6s, 4f, 5d, 6p
Option: 2 6s, 5d, 5f, 6p
Option: 3 6s, 5f, 6d, 6p
Option: 4 6s, 6p, 6d, 6f

Energy order of orbital’s according to Aufbau principle-

The order of orbitals filling is 6s, 4f, 5d, 6p.

Therefore, the correct option is (1).

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Kuldeep Maurya

The difference between the radii of 3rd and 4th orbits of $Li^{2+}$ is $\Delta R_{1}.$ The difference between the radii of 3rd and 4th orbits of $He^{+}$ is $\Delta R_{2}.$ Ratio $\Delta R_{1}:\Delta R_{2}$ is :
Option: 1 8 : 3
Option: 2 3 : 8
Option: 3 2 : 3
Option: 4 3 : 2

We know the formula,

The radius of n^th Bohr's Orbit

$\mathrm{r = 0.529\times\frac{n^2}{Z}}\mathrm{\ A^{\circ}}$

Where
n = principal quantum number of orbit.
Z = Atomic number

Now,

The difference between the radii of 3rd and 4th orbits of $Li^{2+}$ is $\Delta R_{1}.$

$\mathrm{\left(R_{4}-R_{3}\right)_{L i^{+2}}=\frac{0.529}{3}\left\{4^{2}-3^{2}\right\}=\Delta R_{1}}$

The difference between the radii of 3rd and 4th orbits of $He^{+}$ is $\Delta R_{2}.$

$\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)_{\mathrm{He}^{+2}}=\frac{0.529}{2}\left\{4^{2}-3^{2}\right\}=\Delta \mathrm{R}_{2}$

Ratio $\Delta R_{1}:\Delta R_{2}$ is

$\frac{\Delta \mathrm{R}_{1}}{\Delta \mathrm{R}_{2}}=\frac{\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)_{\textup{Li}^{2+}}}{\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)_{\mathrm{He}^{+}}}=\frac{\frac{4^{2}}{3}-\frac{3^{2}}{3}}{\frac{4^{2}}{2}-\frac{3^{2}}{2}}=\frac{7 / 3}{7 / 2}=\frac{2}{3}$

Therefore, the correct option is (3).

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Kuldeep Maurya

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The correct statement about parobability density (except at a distance from nucleus) is :-
Option: 1 it can be zero for 1s orbital
Option: 2 It can never be zero for 2s orbital
Option: 3 It can be negative for 2p orbital
Option: 4 it can be zero for 3p orbital

Probability density can be zero for 3p orbital.

Therefore, Option(4) is correct.

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Kuldeep Maurya

The region in the electromagnetic spectrum where the Balmer series lines appear is:
Option: 1 Visible
Option: 2 Microwave
Option: 3 Infrared
Option: 4 Ultraviolet

Balmer series give visible lines for H-atom.

As lines of Balamer series belongs to both UV as well visible region of EM spectrum. However most appropriate should be visible region.

Therefore, the correct option is (1).

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Kuldeep Maurya

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The shortest wavelength of H atom in the Lyman series is $\lambda_{1}$ . The longest wavelength in the Balmer series He⁺ is:
Option: 1 $\frac{36 \lambda _{1}}{5}$
Option: 2 $\frac{5\lambda _{1}}{9}$
Option: 3 $\frac{9 \lambda _{1}}{5}$
Option: 4 $\frac{27 \lambda _{1}}{5}$

We know,

$\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$

So, $\lambda=\frac{h c}{\Delta E}$ for $\lambda$ minimum(Shortest)

$\mathrm{\Delta \mathrm{E}= maximum}$

for f H atoms in the Lyman series n = 1 & for $\mathrm{\Delta \mathrm{E}_{maximum}}$ Transition must be form n = $\infty$ to n = 1,

Z = 1

$\text { So } \frac{1}{\lambda}=R_{H} Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}} \mathrm{(1)}^{2}(1-0)$

$\frac{1}{\lambda_1}=\mathrm{R_H} \times(1)^{2} \Rightarrow \lambda_{1}=\frac{1}{\mathrm{R_H}}$

For longest wavelength $\mathrm{\Delta \mathrm{E}_{minimum}}$ for Balmer series n = 3 to n = 2 will have $\mathrm{\Delta \mathrm{E}_{minimum}}$ for He⁺ , Z = 2

$\text { So } \frac{1}{\lambda_{2}}=R_{H} \times Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\frac{1}{\lambda_{2}}=\mathrm{R}_{\mathrm{H}} \times (2)^2\left(\frac{1}{4}-\frac{1}{9}\right)$

$\frac{1}{\lambda_{2}}=\mathrm{R}_{\mathrm{H}} \times \frac{5}{9}$

$\lambda_{2}=\frac{1}{R_H} \times \frac{9}{5}$

$\lambda_{2}=\lambda_{1} \times \frac{9}{5}$

Therefore, the correct option is (3).

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Kuldeep Maurya

Consider the hypothetical situation where the azimuthal quantum number, $l,$ takes values $0,1,2,.....n+1,$ where $n$ is the principal quantum number. Then, the element with atomic number :
Option: 1 $9$ is the first alkali metal
Option: 2 $13$ has a half-filled valence subshell
Option: 3 $8$ is the first noble gas
Option: 4 $6$ has a $2p$ -valence subshell

GIven l= 0 to (n + 1)

Now for n = 1,
l = 0, 1, 2

Orbitals :

1s , l =0

1p , l = 1

1d , l = 2

And for

n = 2

l = 0, 1, 2, 3

$\mathrm{Orbitals} : 2s,2p,2d,2f$

And for

n = 3

l = 0, 1, 2, 3,4

$\mathrm{Orbitals} : 3s,3p,3d,3f,3g$

Now, in order to wire electronic configuration, we need to apply (n + l) rule.

Energy order : 1s < 1p < 2s < 1 d < 2p < 3s < 2d ...

1) 9 : 1s² 1p⁶ 2s¹ is the first alkali metal because after losing one electron, it will achieve the first noble gas configuration

2) 13 : 1s² 1p⁶ 2s² 1d³ is not half filled. (Showing incorrect statement)

3) 8 : 1s² 1p⁶ is the first noble gas.

4) 6 : 1s² 1p⁴ has 1p valence subshell.

Therefore, Option 2 is correct.

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Kuldeep Maurya

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All Questions

The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with : Option: 1 Lyman series Option: 2 Balmer series Option: 3 Paschen series Option: 4 Brackett series

Posted by

vishal kumar

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

The number of orbitals associated with quantum numbers is: Option: 1 25 Option: 2 11 Option: 3 15 Option: 4 50

Posted by

Kuldeep Maurya

The radius of the second Bohr orbit, in terms of the Bohr radius, in is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

vishal kumar

JEE Main high-scoring chapters and topics

In the sixth period, the orbitals that are filled are :Option: 1 6s, 4f, 5d, 6pOption: 2 6s, 5d, 5f, 6pOption: 3 6s, 5f, 6d, 6pOption: 4 6s, 6p, 6d, 6f

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The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with :
Option: 1 Lyman series
Option: 2 Balmer series
Option: 3 Paschen series
Option: 4 Brackett series

The number of orbitals associated with quantum numbers $n = 5, m_s = +1/2$
is:
Option: 1 25
Option: 2 11
Option: 3 15
Option: 4 50

The radius of the second Bohr orbit, in terms of the Bohr radius, $a_{0},$ in $Li^{2+}$ is :
Option: 1 $\frac{2a_{0}}{3}$

Option: 2 $\frac{4a_{0}}{3}$

Option: 3 $\frac{4a_{0}}{9}$

Option: 4 $\frac{2a_{0}}{9}$

In the sixth period, the orbitals that are filled are :
Option: 1 6s, 4f, 5d, 6p
Option: 2 6s, 5d, 5f, 6p
Option: 3 6s, 5f, 6d, 6p
Option: 4 6s, 6p, 6d, 6f

The difference between the radii of 3rd and 4th orbits of $Li^{2+}$ is $\Delta R_{1}.$ The difference between the radii of 3rd and 4th orbits of $He^{+}$ is $\Delta R_{2}.$ Ratio $\Delta R_{1}:\Delta R_{2}$ is :
Option: 1 8 : 3
Option: 2 3 : 8
Option: 3 2 : 3
Option: 4 3 : 2

The correct statement about parobability density (except at a distance from nucleus) is :-
Option: 1 it can be zero for 1s orbital
Option: 2 It can never be zero for 2s orbital
Option: 3 It can be negative for 2p orbital
Option: 4 it can be zero for 3p orbital

The region in the electromagnetic spectrum where the Balmer series lines appear is:
Option: 1 Visible
Option: 2 Microwave
Option: 3 Infrared
Option: 4 Ultraviolet

The shortest wavelength of H atom in the Lyman series is $\lambda_{1}$ . The longest wavelength in the Balmer series He⁺ is:
Option: 1 $\frac{36 \lambda _{1}}{5}$
Option: 2 $\frac{5\lambda _{1}}{9}$
Option: 3 $\frac{9 \lambda _{1}}{5}$
Option: 4 $\frac{27 \lambda _{1}}{5}$