[Free] IIT JAM Physics Test Series 2023 : Modern Physics – (Atomic Physics)

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[Free] IIT JAM Physics Test Series 2023 : Modern Physics – (Origin of Quantum Mechanics)

Now In this particular Post of IIT JAM Physics Test Series, you will get a test of the topic Atomic Physics of Chapter Modern Physics. There are total 15 Questions given below also Answers are attached at the end of the test so that you can verify your answers after completing the test. So, Practice these Questions and Do your Best. Also Solve IIT JAM Physics Previous Year Question Paper. And Don’t Forget to Share with Your Friends.


[Free] IIT JAM Physics Test Series 2023 : Modern Physics – (Atomic Physics)


Q1. let Tg and Te be the kinetic energies of the electron in the ground and the third excited states of a hydrogen atom, respectively. According to the Bohr model, the ratio Tg/Te is

(a) 3
(b) 4
(c) 9
(d) 16


Q2. Given that the ionization energies of Hydrogen (1H) and (3Li) are 13.6 eV and 5.39 eV, respectively, the effective nuclear charge experienced by the valence electron of 3Li atom may be estimated in terms of the proton charge e as

(a) 0.63 e
(b) 1.26 e
(c) 1.59 e
(d) 3.00 e


Q3. Consider the high excited states of a Hydrogen atom corresponding to large values of the principal Quantum number (n >> 1). The wavelength λ of a photon emitted due to an electron undergoing a transition between two such states with consecutive values of n (i.e, ψn+1 → ψn ) is related to the wavelength λa of the Ka line of Hydrogen by

(a) λ = n3 λa / 8
(b) λ = 3n3 λa / 8
(c) λ = n2 λa
(d) λ = 4λa /n2


Q4. The velocity of an electron in the ground state of a hydrogen atom is vH . If vp be the velocity of an electron in the ground state of positronium , then

(a) vp = vH
(b) vp = 2vH
(c) vp = vH/2
(d) vp = √2 vH


Q5. A sample of ordinary hydrogen (1H1) gas in a discharge tube was seen to emit the usual Balmer Spectrum. On careful examination , however, it was found that the Ha line in the spectrum was split into two fine lines, one an intense line at 656.28 nm, and the other a faint line at 656.04 nm. From this, one can conclude that the gas sample has a small impurity of

(a) 1H2
(b) 1H3
(c) 2He4
(d) H2O


Q6. The binding energy of the hydrogen atom (electron bound to proton) is 13.6 eV. The binding energy of positronium (electron bound to positron) is

(a) 13.6/180 eV
(b) 13.6/2 eV
(c) 13.6 × 1810 eV
(d) 13.6 × 2 eV


Q7. If a proton were ten times lighter, then the ground state energy of the electron in a Hydrogen atom would have been

(a) Less
(b) More
(c) The same
(d) Depends on the electron mass


Q8. If the Rydberg constant of an atom of finite mass is αR , where R is the Rydberg constant corresponding to an infinite nuclear mass, the ratio of the electronic to nuclear mass of the atom is :

(a) (1 – α)/α
(b) (α – 1)/α
(c) (1 – α)
(d) 1/α


Q9. The wavelength of one of the visible lines in the spectrum of the hydrogen atom is approximately 6560 Å . It is known that this corresponds to a transition between the states with the principal quantum number n = 2 and n = 3. What will be the frequency corresponding to the transition from the ground state to the first excited state

(a) 4.56 × 1014 Hz
(b) 1.65 × 1015 Hz
(c) 2.47 × 1015 Hz
(d) 3.29 × 1015 Hz


Q10. An X-ray machine operates at a potential of 5000 V. What is the minimum possible value of wavelength present in the radiations ?

(a) 0.0124 nm
(b) 0.0245 nm
(c) 0.124 nm
(d) 0.248 nm


Q11. Consider the simple Bohr model of the Hydrogen atom . Its ground state has an energy of – 13.6 eV. The wavelength of the Electromagnetic radiation emitted when the atom makes a transition from first excited state to the ground state is

(a) 1.22 × 10-5 cm
(b) 2.44 × 10-5 cm
(c) 6.22 × 10-5 cm
(d) 4.32 × 10-5 cm


Q12. The first ionization potential of helium atom is 24.6 eV . What is the minimum energy required to remove both the electrons of a neutral helium atom ?

(a) 108.8 eV
(b) 79.0 eV
(c) 49.2 eV
(d) 37.9 eV


Q13. A three-level system of atom has N1 atoms in level E1 ,N2 in Level E2 and N3 in Level E3 (N2 > N1 > N3) and (E1 < E2 < E3) . Laser emission is possible for the transitions :

(a) E2 → E1
(b) E3 → E2
(c) E2 → E3
(d) E3 → E1


Q14. Assuming LS coupling, the lowest energy level for carbon (Z=6) is

(a) 1S0
(b) 3P0
(c) 1D2
(d) 1S1


Q15. Splitting of Spectral lines in the Stark Effect depends on :

(a) Magnetic Quantum Number
(b) Spin Quantum Number
(c) Azimuthal Quantum Number
(d) Principal Quantum Number


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Answer Key (if you find any answer wrong, feel free to Correct us)

01.(d)06.(a)11.(a)
02.(b)07.(b)12.(b)
03.(b)08.(a)13.(a)
04.(a)09.(c)14.(b)
05.(b)10.(d)15.(a)

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