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A hydrogen-like atom (atomic number $Z$ ) is in a higher excited state of quantum number $n$ . This excited atom can make a transition to the first excited state by successively emitting two photons of energies $10.20 eV$ and $17.00 eV$ respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy $4.25 eV$ and $5.95 eV$ respectively. Determine the values of $n$ and $Z$ .

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Important Questions on Atomic Structure

EASY

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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:

EASY

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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

EASY

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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

HARD

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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>Atomic Structure>Bohr’s model of atom

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

MEDIUM

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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}

MEDIUM

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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>Atomic Structure>Bohr’s model of atom

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

MEDIUM

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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>Atomic Structure>Bohr’s model of atom

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>Atomic Structure>Bohr’s model of atom

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

MEDIUM

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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>Atomic Structure>Bohr’s model of atom

The longest wavelength of Balmer series of

H

atom is

6563 \overset{o}{A}

. Then the longest wavelength of Lyman series is

MEDIUM

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

If the shortest wavelength in Lyman series of hydrogen atom is

A

, then the longest wavelength in Paschen series of

{He}^{+}

EASY

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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

EASY

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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

MEDIUM

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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

EASY

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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

MEDIUM

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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 :

EASY

Chemistry>Physical Chemistry>Atomic Structure>Bohr’s model of atom

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}

Important Questions on Atomic Structure

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