Test: Atomic & Molecular Physics - 1 - Physics MCQ

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 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 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 Zeeman splitting ....(i)
∵ 
∴ Equation (i) ⇒ 
⇒ 

Tlie commom wave number difference in the two successive rotational lives 

The difference of energy between energy levels having principal quantum number n and (n + 1) is given by

The Lande’s g-factor is defined as
...(i)
For ³P₂, S = 1, L = 1, J = 2
Therefore, equation (i) ⇒

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.

Since, 
For λ = 600 nm, the line width is given by

The excitation potential is given by



= 4.9 volt.

Since, 
 , where I is the moment of inertia of the rotating molecule.
Therefore, 

From Moseley's law to r Kα - line, we have 
Therefore, 
Therefore, 

The total angular momentum 
Taking self dot product on both sides, we get 
⇒ 
⇒ 
⇒ 

For posittronium atom reduced mass,

The Rydbery constant for positronium atom, R_P 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 (n_i = 3 to n_f = 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

⇒ 
⇒ .....(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 

For ²S_1/2 electronic level: 
For gBe⁺ ion, I = 3/2
The hyperfine quantum number F is defined as
 with decrease o f unity.

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

Since, doublet splitting of a one election atomic state arising due to spin orbit interaction is given by

Therefore, 
Therefore, 

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 (E_diss), we have

⇒
⇒ 
⇒  

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