Choose the correct nuclear process from the below options [p: proton, n: neutron, mathbfe^- : electron, mathrme^+ : positron, v: neutrino, overlinenu : antineutrino]

Solution & Explanation

### Core Logic In basic beta^- emission processes, a neutron decays inside a nucleus to satisfy lepton numbers and conservation rules: mathrmn rightarrow mathrmp + mathrme^- + overlinenu ### Step 1: Conservation Cross-Check Charge Balance: 0 rightarrow (+1) + (-1) + 0 = 0 (Conserved) Lepton Family Index: 0 rightarrow 0 + (+1) + (-1) = 0 (Conserved via antineutrino entry). This perfectly isolates option (1). ### Pattern Recognition Negative beta emission is always accompanied by an antineutrino, whereas positive positron transformation releases a regular neutrino molecule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

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Q60 jee_main_2024_31_jan_evening Nuclear Size and Density
A nucleus has mass number A_1 and volume V_1. Another nucleus has mass number A_2 and volume V_2. If relation between mass number is A_2 = 4A_1, then fracV_2V_1 = ________.
Numerical Answer. Answer: 4 to 4

Solution

### Related Formula R = R_0 A^1/3 V = frac43pi R^3 ### Core Logic Since radius R is proportional to A^1/3, the volume V (which depends on R^3) will be directly proportional to the mass number A. ### Step 1: Show Proportionality V = frac43pi (R_0 A^1/3)^3 = frac43pi R_0^3 A This proves that V propto A. ### Step 2: Calculate Ratio fracV_2V_1 = fracA_2A_1 Given that A_2 = 4A_1: fracV_2V_1 = frac4A_1A_1 = 4 ### Pattern Recognition Nuclear density is constant for all nuclei. Therefore, Mass propto Volume. Since Mass number (A) represents mass, Volume is strictly linearly proportional to Mass number. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei
Q60 jee_main_2024_31_jan_morning Mass Defect And Energy
The mass defect in a particular reaction is 0.4mathrm\ g. The amount of energy liberated is n times 10^7mathrm\ kWh where n = _______. (speed of light = 3 times 10^8mathrm\ m/s)
Numerical Answer. Answer: 1 to 1

Solution

### Related Formula E = Delta m c^2 1 text kWh = 3.6 times 10^6 text J ### Core Logic Given the mass defect: Delta m = 0.4mathrm\,g = 0.4 times 10^-3mathrm\,kg The total energy liberated in Joules is: E = (0.4 times 10^-3) times (3 times 10^8)^2 E = 0.4 times 10^-3 times 9 times 10^16 E = 3.6 times 10^13mathrm\,J ### Step 2: Conversion to kWh We need the answer in mathrmkWh. Since 1mathrm\,kWh = 1000mathrm\,W times 3600mathrm\,s = 3.6 times 10^6mathrm\,J: E = frac3.6 times 10^133.6 times 10^6mathrm\,kWh E = 10^7mathrm\,kWh Comparing this to n times 10^7mathrm\,kWh, we get: n = 1 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

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