Skip to main content
Back to ATOMS

Your Progress

0 / 130 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • ATOMS

Atoms practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 121-130 of 130 questions

Preparing practice and saved progress…

121 Easy
121 of 130
Application +4 | -1
An electron revolving in nthn^{\text {th}} Bohr orbit has magnetic moment μn.\mu_n. If μn∝nx,\mu_n \propto n^x, the value of xx is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the magnetic moment is directly proportional to the principal quantum number n.
122 Medium
122 of 130
Analyzing +4 | -1
v1{v_1} is the frequency of the series limit of Lyman series,v2{v_2} is the frequency of the first line of Lyman series and v3{v_3} is the frequency of the series limit of the Balmer series. Then
View answer and explanation
Correct option: A
v1−v2=v3{v_1} - {v_2} = {v_3}
Option A (v1−v2=v3{v_1} - {v_2} = {v_3}) is correct because the frequency of the series limit of the Lyman series minus the frequency of the first line of the Lyman series equals the frequency of the series limit of the Balmer series.
123 Easy
123 of 130
Application +4 | -1
A electron collides elastically with HH-like atom and excites it from ground state to n=3n=3. Find out the energy transfer to HH-like atom.
View answer and explanation
Correct option: A
12.1 eV12.1 \mathrm{~eV}
Option A (12.1 eV12.1 \mathrm{~eV}) is correct because the energy transfer is equal to the energy difference between the ground state and the n=3 state, which is 12.1 eV for a hydrogen-like atom with Z=3.
124 Medium
124 of 130
Application +4 | -1
An Electron is revolving around nucleus of charge Ze, if 40.8 eV of energy needed to excite an electron from n=1 to n=2 state. What is value of Z?
View answer and explanation
Correct option: A
2
Option A (2) is correct because the energy difference between n=1 and n=2 is given by ΔE=13.6Z2(112−122)\Delta E = 13.6 Z^2 (\frac{1}{1^2} - \frac{1}{2^2}). Solving for Z gives Z=2.
125 Easy
125 of 130
Application +4 | -1
Energy of electron (in eV\mathrm{eV}) in 2nd2^{\text {nd}} orbit of He+He^+ ion is
View answer and explanation
Correct option: B
−13.6 eV-13.6 \mathrm{~eV}
Option B (−13.6 eV-13.6 \mathrm{~eV}) is correct because the energy of an electron in the nth orbit of a hydrogen-like atom is given by En=−13.6Z2n2eVE_n = -13.6 \frac{{Z^2}}{{n^2}} eV. For He+He^+, Z=2 and n=2, so E2=−13.62222=−13.6eVE_2 = -13.6 \frac{{2^2}}{{2^2}} = -13.6 eV.
126 Medium
126 of 130
Remembering +4 | -1
The significant result deduced from the
Rutherford's scattering experiment is that
View answer and explanation
Correct option: A
whole of the positive charge is concentrated at
the centre of atom
Option A (whole of the positive charge is concentrated at the centre of atom) is correct because Rutherford's experiment showed that most alpha particles passed through the gold foil undeflected, but a few were deflected at large angles, indicating a small, dense, positively charged nucleus.
127 Medium
127 of 130
Application +4 | -1
The ratio of the wavelengths for 2→12 \to 1 transition on Li++,He+L{i^{ + + }},H{e^ + } and HH is

View answer and explanation
Correct option: C
4:9:36\mathrm{4 : 9 : 36}
Option C (4:9:36\mathrm{4 : 9 : 36}) is correct because the wavelength is inversely proportional to Z2, where Z is the atomic number.
128 Hard
128 of 130
Analyzing +4 | -1
An XX-rays tube is working at a potential difference of 38.08 kV38.08 \,\mathrm{kV}. The potential difference is decreased to half its initial value. It is found that difference of the wavelength of Kα XK_\alpha \,{X}-ray and the most energetic continuous XX-rays becomes 4 times of the difference prior to the change of voltage. Assuming KαK_\alpha line is present in both cases, find the atomic number of the target element. [Take: Rch=13.6 eVR c h=13.6 \,\mathrm{eV}]
View answer and explanation
Correct option: A
41
Option D ( almost remain same) is correct because the nucleus density is independent of the mass number.
129 Hard
129 of 130
Application +4 | -1
In the Bohr model of a π\pi-mesic atom, a π\pi-meson of mass mπm_\pi and of the same change as the electron is in a circular orbit of radius rr about the nucleus with an orbital angular momentum h2π\frac{h}{2 \pi}. If the radius of a nucleus of atomic number ZZ is given by R=1.6×10−15Z13 mR=1.6 \times 10^{-15} Z^{\frac{1}{3}} \mathrm{~m}. The limit on ZZ for which (ε0h2πme2=0.53A∘\left(\frac{\varepsilon_0 h^2}{\pi m e^2}=0.53 \mathop A\limits^ \circ\right. and mπme=264)π\left.\frac{m_{\pi}}{m_e}=264\right) \pi-mesic atoms might exist is:
View answer and explanation
Correct option: C
<37< 37
Option C (< 37) is correct because the radius of the mesic atom must be greater than the radius of the nucleus for it to exist, leading to the condition Z < 37.
130 Medium
130 of 130
Understanding +4 | -1
Which of the relation is correct between time period and number of orbits while an electron is revolving in an orbit
View answer and explanation
Correct option: C
n3{n^3}
Option C (n3{n^3}) is correct because the time period of an electron revolving in an orbit is proportional to n3{n^3}, where n is the principal quantum number.
Revision notes and key takeaways

Review ATOMS in your textbook or class notes, then use this set to check your understanding.

Showing 121-130 of 130 questions

Your practice

0 / 130 answered · 130 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.