EASY
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Write down the wavelength of the last line of Balmer series in Angstrom.

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Important Questions on Atoms

EASY
If λ1 and λ2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of λ1:λ2 is :
EASY
In Balmer series for hydrogen atom, find the energy of photon corresponding to longest wavelength.
HARD
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
EASY
The wavelength of the first spectral line of the Lyman series of hydrogen spectrum is
EASY
Given the value of Rydberg constant is 107 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be:
MEDIUM
The recoil speed of a hydrogen atom after it emits a photon in going from n=5 state to n=5 state will be
MEDIUM
The shortest wavelength of Paschen series in hydrogen spectrum is 8182 A. The first member of the Paschen series is nearly
EASY
Calculate the minimum wavelength of the spectral line present in Balmer series of hydrogen.
MEDIUM
Taking the wavelength of first Balmer line in hydrogen spectrum (n=3  to n=2) as 660 nm , the wavelength of the 2nd Balmer line (n = 4  to  n = 2) will be :
EASY
In Balmer series, wavelength of first line is λ1 and in Brackett series wavelength of first line is λ2 then λ1λ2 is
EASY
The frequency for a series limit of Balmer and Paschen series respectively are f1 and f3. If the frequency of the first line of Balmer series is f2 then the relation between f1, f2 and f3 is
MEDIUM
If the series limit frequency of the Lyman series is VL, then the series limit frequency of the Pfund series is:
MEDIUM
The longest wavelength of Balmer series of H atom is 6563 Ao. Then the longest wavelength of Lyman series is
EASY
The ratio of wavelengths of the last line of the Balmer series and the last line of the Lyman series is_____.
HARD
Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom?
MEDIUM

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

EASY
Name the series of hydrogen spectrum lying in ultraviolet and visible region.
EASY
When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the
MEDIUM
The first member of the Balmer series of hydrogen atom has a wavelength of 6561 A. The wavelength of the second member of the Balmer series (in nm) is_____________
EASY
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is: