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The first five series of lines that correspond to $n_{1} = 1, 2, 3, 4, 5$ are known as Lyman, Balmer, Paschen, and Pfund series, respectively.

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Important Questions on Structure of Atom

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

The shortest wavelength of

H

atom in the Lyman series is

λ_{1}

. The longest wavelength in the Balmer series of

{He}^{+}

is :

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about

9

. The spectral series are:

EASY

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the

HARD

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

In a muonic atom, a muon of mass of

200

times of that of electron and same charge is bound to the proton. The wavelengths of its Balmer series are in the range of

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

If the shortest wavelength in Lyman series of hydrogen atom is

A

, then the longest wavelength in Paschen series of

{He}^{+}

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

If the wave number of first spectral line of Lyman series of

{He}^{+}

{Xcm}^{- 1}

, then the wave number of second line of Balmer series of hydrogen atom in

{cm}^{- 1}

EASY

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

In Balmer series, wavelength of first line is

λ_{1}

and in Brackett series wavelength of first line is

λ_{2}

then

\frac{λ_{1}}{λ_{2}}

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

Draw the energy level diagram of the hydrogen atom and show transitions responsible for emission lines of the Balmer series.

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

The first member of the Balmer series of hydrogen atom has a wavelength of

6561 \overset{\circ}{A}

. The wavelength of the second member of the Balmer series (in nm) is_____________

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

Taking the wavelength of first Balmer line in hydrogen spectrum

(n = 3

n = 2)

660 n m

, the wavelength of the

2^{n d}

Balmer line

(n = 4

n = 2)

will be :

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

Identify the correct statements regarding Balmer series: The ware number of each line in the spectrum of atomic hydrogen is given by the equation

\bar{v} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) .

where

R_{H}

is a constant and

n_{1}

and

n_{2}

are integers.
(i) As ware length decreases, the lines in the series converge
(ii) The integer

n_{1}

is equal to

2

(iii) The ionization energy of hydrogen can be calculated from the wave number of these lines
(iv) The line of longest wavelength corresponds to

n_{2} = 3

EASY

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

In Balmer series for hydrogen atom, find the energy of photon corresponding to longest wavelength.

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

The longest wavelength of Balmer series of

H

atom is

6563 \overset{o}{A}

. Then the longest wavelength of Lyman series is

EASY

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

Two series of spectral lines of atomic hydrogen which do not belong to infrared spectral region are

EASY

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

Calculate the minimum wavelength of the spectral line present in Balmer series of hydrogen.

EASY

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

Find the ratio of maximum wavelength to minimum wavelength for the lines of Balmer series in hydrogen spectrum.

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

Which spectral series of hydrogen spectrum lies in the visible region of electromagnetic spectrum?

MEDIUM

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

For any given series of spectral lines of atomic hydrogen, let $Δ \bar{v} = {\bar{v}}_{m a x} - {\bar{v}}_{m i n}$ be the difference in maximum and minimum wave number in $c m^{- 1}$ .

The ratio $Δ {\bar{v}}_{L y m a n} / Δ {\bar{v}}_{B a l m a r}$ is

EASY

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

The frequency for a series limit of Balmer and Paschen series respectively are

f_{1}

and

f_{3}

. If the frequency of the first line of Balmer series is

f_{2}

then the relation between

f_{1}, f_{2}

and

f_{3}

EASY

Chemistry>Physical Chemistry>Structure of Atom>Atomic Spectra

Given the value of Rydberg constant is

10^{7} m^{- 1}

, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

The first five series of lines that correspond to n1=1,2,3,4,5 are known as Lyman, Balmer, Paschen, and Pfund series, respectively.

Important Questions on Structure of Atom

Learn and Prepare for any exam you want !

The first five series of lines that correspond to $n_{1} = 1, 2, 3, 4, 5$ are known as Lyman, Balmer, Paschen, and Pfund series, respectively.