The radius of the first Bohr orbit in for a μ-mesic (or muonic) atom is ( The masses of μ– -meson and proton are 207 times and 1836 times respectively the mass of electron).
The radius of the first orbit (n = 1) of a hydrogen like muonic atomtaking finite mass ofnucleus in consideration is
....(i) where μ is reduced mass.
The reduced mass of the muonic atom (formed by a μ-meson and a proton) is
, where m is the mass o f the electron.
Thus. equation (i) ⇒
The moment of inertia of the HCl molecule is 2.71 x 10-47 kg-m2. The most populated rotational level for the molecule at temperature of 600 K corresponds to
The value of rotational quantum number J tor the maximum populated rotational level for a molecule at a temperature T is given by J =, where B is rotational constant.
∴
= 4.5 - 0.5 =4.
The magnetic moment of an election moving in a circular orbit of radius r about a proton is
where =
The magnetic moment is given by
...(i)
For electron the required centripetal force comes from Coulombic attraction between nucleus and electron.
...(ii)
From equation (ii), equation (i) ⇒
The normal Zeeman splitting in the mercury 4916 Å line in a magnetic field of 0.3 T is
The Zeeman splitting ....(i)
∵
∴ Equation (i) ⇒
⇒
The common wave number difference in the two successive rotational lines is
Tlie commom wave number difference in the two successive rotational lives
For high principal quantum number (n) for hydrogen atom the spacing between the neighbouring energy level is proportional to
The difference of energy between energy levels having principal quantum number n and (n + 1) is given by
The Lande g-factor for the 3P2 level of an atom is______________
(Round off to one decimal places)
The Lande’s g-factor is defined as
...(i)
For 3P2, S = 1, L = 1, J = 2
Therefore, equation (i) ⇒
The Bohr model gives the value for the ionization potential of Be+++ ion as__________10-1 eV
(Round off to two decimal places)
The ionization energy of Be+++ = Z2 x ionization energy of hydrogen atom
= (4)2x 13.6 e V = 16 x 13.6 eV = 217.6 eV.
The line width of a spectral line of 600 mil wavelength emitted from a level of a lifetime of 10-8 sec, is__________10-15 m (up to two decimal places)
Since,
For λ = 600 nm, the line width is given by
The radiations given offby the Hg atoms returning to their normal states were studied by Frank and Heitz and
a line was observed at 2573 Å. The excitation potential for Hg is________volt (upto one decimal place)
The excitation potential is given by
= 4.9 volt.
The spacing of a series of lines in the microwave spectrum of AIH is constant at 12.604 cm-1. The moment of
inertia of the AIH molecule is_______x 10-47 kg/m2. (upto two decimal places)
Since,
, where I is the moment of inertia of the rotating molecule.
Therefore,
If the Kα -radiation o f Mo (Z = 42) lias a wavelength o f 0.71 A. the wavelength o f the corresponding radiation for Cu (Z = 29) is Å. (upto two decimal places)
From Moseley's law to r Kα - line, we have
Therefore,
Therefore,
The possible values of is
The total angular momentum
Taking self dot product on both sides, we get
⇒
⇒
⇒
A positroniiun atom is system consisting of a position and an electron revolving about their common center of mass, which lies halfway between them. The ratio of the wavelength of the first line of the Balmer series of positroniumatomto the wavelength of the first line of the Balmer seiies of hydrogen atom is
For posittronium atom reduced mass,
The Rydbery constant for positronium atom, RP is given by
...(1)
Now, the wavelength of Balmer first line of positronium a tom is given by
....(2)
Similarly, for hydrogen atom, the wavelength of Balmer first line (ni = 3 to nf = 2) is given by
Now, the ratio of wavelengths of first line o f Balmer series o fpositronium a tom and hydrogen a tom is
A sample of ordinary hydrogen gas in a discharge tube was seen to emit the usual Balmer spectrum. On careful examination however, it was found that the Hα -line in the spectrum was split into two fine lines, one an intense line at 6562.80 Å and the other at 6561.01 Å. From this, one can conclude that the gas sample had a small impurity of
⇒
⇒ .....(i)
For the
-line is at 6562.80 Å. thus equation (i) implies with M = 1836 m, λ = 6562.80 Å;
...(ii)
For unknown impurity, the - line is at 6561.01 Å thus M = x x 1836 m (H ere, we have assum ed the m ass of nucleus of unknown impurity equal to x times mass of proton or hydrogen nucleus). λ = 6561.01 Å;
...(iii)
Now. dividing equation (ii) by (iii), we get 1.000272824
⇒
⇒
Hence, the mass o f nucleus is 2 times that of Out o f the given options, nucleus
has the mass equal to two times the nuclear mass of
The gBe+ ion has a nucleus with spin The possible values of the hypofine quantum number F for the 2S1/2 electric level are
For 2S1/2 electronic level:
For gBe+ ion, I = 3/2
The hyperfine quantum number F is defined as
with decrease o f unity.
An atomic magnetic dipole having orbital magnetic moment of 1 Bohr magneton is aligned parallel to a external magnetic field of 1 Tesla. The energy required to turn the dipole to align it antiparallel to the field is_________x 10-4 eV. (upto two decimal places)
The potential energy of dipole
For parallel alignment, θ = 0°
...(i)
For antiparallel alignment, θ = 180°
...(ii)
The energy required to align the dipole in antiparallel direction to the field
In the doublet splitting of the first excited state. 22P3/2 — 2 2P1/2 of He+ is 5.84 cm-1. the corresponding separation for H is
Since, doublet splitting of a one election atomic state arising due to spin orbit interaction is given by
Therefore,
Therefore,
Assume that the H2 molecule behaves like a harmonic oscillator with a force constant k = 573 N/m the vibrational quantum number, corresponding to its dissociation energy 4.5 eV, is
For H2, the reduced mass (μ)
Therefore, frequency of oscillation of the molecule is given by
For vibrational quantum number (n) corresponding to dissociation energy (Ediss), we have
⇒
⇒
⇒
If the exciting line in an experiment is 5460 Å and the stokes line is at 5520 Å. the wavelength (inÅ) of the antistokes line is__________Å. (should be an integer)
Since, the stokes and the anti-stokes lines have the same wave number difference with respect to the exciting line. The w ave num ber o f the exciting line is v = and that of the stokes line is
Thus, the wave num ber displacement is
Therefore, the wave number corresponding to the antistokes line would be given by
The corresponding wavelength is
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