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Given below are two statements: Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model. Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model. In the light of the above statements, choose the most appropriate from the options given below:

Solution & Explanation

### Core Logic * **Statement I**: According to Rutherford's planetary model of the atom, almost the entire mass of an atom and all of its positive charge are concentrated in a centrally located small volume called the nucleus, with electrons revolving around it. Thus, Statement I is **true**. * **Statement II**: Thomson's model of the atom (plum-pudding model) describes the atom as a spherical cloud of positive charge with electrons embedded in it. This model was proposed before Rutherford's nuclear model and is not a special case of Rutherford's model. Thus, Statement II is **false**. ### Step 1: Conclusion Therefore, Statement I is true but Statement II is false. ### Pattern Recognition Understand the evolutionary timeline of atomic models: 1. Thomson's Plum Pudding Model (spherical cloud with embedded electrons) 2. Rutherford's Nuclear Model (planetary orbital model with tiny centralized nucleus) They are conceptually distinct, making Statement II false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Reference Study Guides

Atoms

More Atoms Previous-Year Questions — Page 3

Q53 jee_main_2024_29_january_evening Bohr's Model of Hydrogen Atom and Hydrogen Spectrum
Hydrogen atom is bombarded with electrons accelerated through a potential difference of V, which causes excitation of hydrogen atoms. If the experiment is being performed at T = 0text K, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be fracalpha10text V, where alpha = ________.
Numerical Answer. Answer: 121 to 121

Solution

### Related Formula The energy of an electron in the n-th state of a hydrogen atom is: E_n = -frac13.6n^2text eV ### Core Logic At T = 0text K, all hydrogen atoms are in their ground state (n = 1). To observe any lines in the Balmer series emission spectrum: * The emission lines must result from a transition terminating in the state n = 2. * This requires the hydrogen electron to first be excited to at least the state n = 3. Thus, the bombarding electrons must have enough kinetic energy to excite the ground-state electron (n = 1) to the n = 3 state. ### Step 1: Calculate the Required Energy and Potential The required energy difference is: Delta E = E_3 - E_1 = -frac13.63^2 - left(-frac13.61^2right)text eV Delta E = 13.6 left(1 - frac19right)text eV = 13.6 times frac89text eV approx 12.09text eV Since the excitation energy is provided by electrons accelerated through potential V, the minimum potential difference required is: V approx 12.09text V approx 12.1text V ### Step 2: Solve for Alpha Comparing to the given form fracalpha10text V: 12.1 = fracalpha10 implies alpha = 121 ### Pattern Recognition Balmer series emission requires excitation of the electron to at least n=3. Ground state n=1 to n=3 excitation energy is 12.09text eV approx 12.1text eV, matching alpha = 121 when multiplied by 10. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms
Q41 jee_main_2024_27_jan_morning Bohr Model of Hydrogen Atom
The radius of the third stationary orbit of an electron for a Bohr's atom is R. The radius of the fourth stationary orbit will be:
  • A. frac43R
  • B. frac169R
  • C. frac34R
  • D. frac916R

Solution

### Related Formula r_n propto fracn^2Z Where n is the principal quantum number and Z is the atomic number. ### Core Logic For a given atom, Z is constant, hence: fracr_4r_3 = left(frac43right)^2 = frac169 Given r_3 = R: ### Step 1: Calculate the final radius r_4 = frac169R ### Pattern Recognition Bohr orbital dimensions scale quadratically with quantum index (n^2), rendering quick ratios straightforward via simple squaring operations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms
Q51 jee_main_2024_29_jan_morning Bohr Model of Hydrogen Atom
When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is fracx5 mathrm~m / s. Where x = _______. (Use: mass of hydrogen atom = 1.6 times 10^-27 mathrm~kg)
Numerical Answer. Answer: 17 to 17

Solution

### Related Formula By conservation of linear momentum, the momentum of the recoiling hydrogen atom matches the momentum of the emitted photon: p_textatom = p_textphoton implies m v = fracDelta Ec Hence, the recoil speed v is: v = fracDelta Em c where Delta E is the energy difference between the electronic energy levels. ### Core Logic For a hydrogen transition from n = 2 to n = 1: Delta E = E_2 - E_1 = -3.4 mathrm~eV - (-13.6 mathrm~eV) = 10.2 mathrm~eV Converting energy to Joules: Delta E = 10.2 times 1.6 times 10^-19 mathrm~J
Electronic transition energy level diagram from n=2 to n=1 for Q51
Electronic transition energy level diagram from n=2 to n=1 for Q51
### Step 1: Compute Recoil Speed Substituting given constant values: v = frac10.2 times 1.6 times 10^-19 mathrm~J(1.6 times 10^-27 mathrm~kg) times (3 times 10^8 mathrm~m/s) ### Step 2: Simplify Arithmetic Fractions Cancelling out the factor 1.6 from numerator and denominator: v = frac10.2 times 10^-193 times 10^-19 = frac10.23 = 3.4 mathrm~m/s ### Step 3: Solve for Variable x Equating the computed speed to the given parameter form: 3.4 = fracx5 implies x = 3.4 times 5 = 17 Therefore, the value of x is 17. ### Pattern Recognition Always look for clean component cancellation matches prior to raw exponential expansion. In this problem, the value 1.6 in the electron charge value cancels with the structural atomic mass constant value perfectly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms
Q37 jee_main_2024_30_january_evening Bohr Model Magnetic Moment
An electron revolving in n^textth Bohr orbit has magnetic moment mu_mathrmn. If mu_mathrmn propto n^mathrmx, the value of mathbfx is:
  • A. 2
  • B. 1
  • C. 3
  • D. 0

Solution

### Related Formula mu_n = i A = fraceT pi r^2 Alternative relation using angular momentum: mu_n = frace2m L ### Core Logic From Bohr's quantization condition, angular momentum L is given by: L = fracnh2pi Therefore, the magnetic moment can be written directly as: mu_n = frace2m left(fracnh2piright) implies mu_n propto n ### Step 1: Direct Proportionality Analysis We can also derive this via velocity and radius: v propto frac1n and r propto n^2. mu_n = frace v r2 propto left(frac1nright) (n^2) = n Comparing with mu_n propto n^x, we find x = 1. ### Pattern Recognition Magnetic moment of an electron in a Bohr orbit is always directly proportional to the principal quantum number n. It is an integer multiple of the Bohr magneton (mu_B = fraceh4pi m). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms
Q36 jee_main_2024_30_jan_morning Bohr Model and Electron Energy
The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5^textth excited state of a hydrogen atom is :
  • A. 4
  • B. frac14
  • C. frac12
  • D. 1

Solution

### Related Formula textKE = -E textPE = 2E textKE = frac12 |textPE| ### Core Logic In any allowed Bohr orbit (for any value of principal quantum number n), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field. |textPE| = 2 times textKE ### Step 1: Formulate the Ratio The question asks for the ratio of the magnitude of KE to the magnitude of PE. fractextKE|textPE| = frac12 This ratio is independent of the orbit state. Even though it is the 5^textth excited state, the ratio remains frac12. ### Pattern Recognition Energy relationships in Bohr orbits (and planetary motion): K = -E = -U/2. The state number (n) is given solely as a distractor. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

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