Advertisements
Advertisements
प्रश्न
Calculate the wavelength of the first two lines in the Balmer series of hydrogen atoms.
उत्तर
For Balmer series,
n = 2, m = 3, 4, 5, ....
(a) For first line in Balmer series, n = 2, m = 3
`1/lambda_1 = R_H(1/n^2 - 1/m^2)`
= `R_H (1/(2)^2 - 1/(3)^3)`
= `R_H (1/4 - 1/9)`
∴ `1/lambda_1 = R_H5/36`
∴ `lambda_1 = 36/(5 R_H)`
Putting RH = 1.097 × 107 m−1
∴ `lambda = 36/(5 xx 1.097 xx 10^7)`
= `(36 xx 10^-7)/5.485`
= 6.563 × 10−7 m
= 6563 Å
(b) For the second line in Balmer series, n = 2, m = 4
∴ `1/lambda_2 = R_H(1/(2)^2 - 1/(4)^2)`
= `R_H (1/4 - 1/16)`
∴ `1/lambda_2 = R_H3/16`
∴ `lambda_2 = 16/(3 R_H)`
= `16/(3 xx 1.097 xx 10^7)`
= `16/3.291 xx 10^-7`
= 4.862 × 10−7 m
= 4862 Å
APPEARS IN
संबंधित प्रश्न
An electron jumps from fourth to first orbit in an atom. How many maximum number of spectral lines can be emitted by the atom? To which series these lines correspond?
In both β− and β+ decay processes, the mass number of a nucleus remains the same, whereas the atomic number Z increases by one in β− decay and decreases by one in β+ decay. Explain giving reason.
Determine the series limit of Balmer, Paschen, and Pfund series, given the limit for Lyman series is 912 Å.
In Balmer series, wavelength of first line is 'λ1' and in Brackett series wavelength of first line is 'λ2' then `lambda_1/lambda_2` is ______.
Which of the following transition will have highest emission wavelength?
Which of the following is TRUE?
If wavelength for a wave is `lambda = 6000 Å,` then wave number will be ____________.
Let the series limit for Balmer series be 'λ1' and the longest wavelength for Brackett series be 'λ2'. Then λ1 and λ2 are related as ______.
The energy (in eV) required to excite an electron from n = 2 to n = 4 state in hydrogen atom is ____________.
Let v1 and v3 be the frequency for series limit of Balmer and Paschen series respectively. If the frequency of first line of Balmer series is v2 then, relation between v1 and v2 and v3 is ____________.
In hydrogen spectrum, which of the following spectral series lies in ultraviolet region?
In hydrogen spectrum, the series of lines obtained in the ultraviolet region of the spectrum is ____________.
If the mass of the electron is reduced to half, the Rydberg constant ______.
In hydrogen spectrum, the wavelengths of light emitted in a series of spectral lines is given by the equation, `1/lambda` = R `(1/4^2 - 1/"n"^2)`, where n = 5, 6, 7...... and 'R' is Rydberg's constant. Identify the series and wavelength region.
Each element is associated with a ______.
Continuous spectrum is produced by ______.
An electron makes a transition from orbit n = 4 to the orbit n = 2 of a hydrogen atom. What is the wave number of the emitted radiations? (R = Rydberg's constant)
What is the minimum energy that must be given to a H atom in ground state so that it can emit an Hγ line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such Hγ photon?
The first four spectral lines in the Lyman series of a H-atom are λ = 1218 Å, 1028Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.
Determine the series limit of Balmer, Paschen and Brackett series, given the limit for Lyman series is 911.6 Å.
The frequencies for series limit of Balmer and Paschen series respectively are 'v1' and 'v3'. If frequency of first line of Balmer series is 'v2' then the relation between 'v1', 'v2' and 'v3' is ______.
In the hydrogen atoms, the transition from the state n = 6 to n = 1 results in ultraviolet radiation. Infrared radiation will be obtained in the transition.
The frequency of the series limit of the Balmer series of the hydrogen atoms of Rydberg’s constant R and velocity of light c is ______.
Find the wavelength and wave number of the first member of the Balmer series in Hydrogen spectrum. (`R =1.097xx10^7m^(-1)`)
