Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let µH and µD be the reduced masses of electrons for hydrogen and deuterium respectively.
We know that `1/λ = R[1/n_f^2 - 1/n_1^2]`
As ni and nf are fixed for by mass series for hydrogen and deuterium.
`λ ∝ 1/R` or `λ_D/λ_H = R_H/R_D` ......(i)
`R_R = (m_ee^4)/(8ε_0ch^3) = (µ_He^4)/(8ε_0ch^3)`
`R_D = (m_ee^4)/(8ε_0ch^3) = (µ_De^4)/(8ε_0ch^3)`
∴ `R_H/R_D = µ_H/µ_D` ......(ii)
From equations (i) and (ii)
`λ_D/λ_H = µ_H/µ_D` ......(iii)
Reduced mass for hydrogen,
`µ_H = m_e/(1 + m_e/M) ≃ m_e(1 - m_e/M)`
Reduced mass for deuterium,
`µ_D = (2M * m_e)/(2M(1 + m_e/(2M))) ≃ m_e(1 - m_e/M)`
Where M is the mass of proton
`µ_H/µ_D = (m_e(1 - m_e/(2M)))/(m_e(1 - m_e/(2M))) = (1 - m_e/M)(1 - m_e/(2M))^-1`
= `(1 - m_e/M)(1 + m_e/(2M))`
⇒ `µ_H/µ_D = (1 - m_e/(2M))`
or `µ_H/µ_D = (1 - 1/(2 xx 1840))` = 0.99973 .....(iv) (∵ M = 1840 me)
From (iii) and (iv)
`λ_D/λ_H` = 0.99973, λD = 0.99973 λH
Using λH = 1218 Å, 1028 Å, 974.3 Å and 951.4 Å, we get
λD = 1217.7 Å, 1027.7 Å, 974.04 Å, 951.1 Å
Shift in wavelength (λH – λD) = 0.3 Å.
APPEARS IN
संबंधित प्रश्न
In the figure below, D and E respectively represent
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 ______.
Which of the following transition will have highest emission frequency?
In hydrogen atom, the product of the angular momentum and the linear momentum of the electron is proportional to (n = principal quantum number) ____________.
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.
Continuous spectrum is produced by ______.
Absorption line spectrum is obtained ______.
To produce an emission spectrum of hydrogen it needs to be ______.
Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in 1H and 2H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass µ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here µ = meM/(me + M) where M is the nuclear mass and m e is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in 1H and 2H. (Mass of 1H nucleus is 1.6725 × 10–27 kg, Mass of 2H nucleus is 3.3374 × 10–27 kg, Mass of electron = 9.109 × 10–31 kg.)
Determine the shortest wavelengths of Balmer and Pasch en series. Given the limit for the Lyman series is 912 Å.
Determine the series limit of Balmer, Paschen and Brackett series, given the limit for Lyman series is 911.6 Å.
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 ______.
