356348
Assertion : Large angle of scattering of alpha particles led to the discovery of atomic nucleus. Reason : Entire positive charge of atom is concentrated in the central core.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\alpha - \) particle is positively charged, so is the nucleus, so the large angle of scattering of \(\alpha - \) particle shows that the nucleus is positively charged and concentrated in the central core. So option (1) is correct.
PHXII12:ATOMS
356350
The trajectory of an \(\alpha\)-particle can be computed by employing:
1 Newton's second law of motion.
2 Coulomb's law for electrostatic force of repulsion between the \(\alpha\)-particle and the positively charges nucleus.
3 Both of these.
4 None of these.
Explanation:
For situation at sub atomic level also basic laws are applicable.
PHXII12:ATOMS
356351
When an \(\alpha - \) particle of mass \('m'\) moving with velocity \('v'\) bombards on a heavy nucleus of charge \('Ze'\), its distance of closest approach from the nucleus depends on \(m\) as:
1 \(\frac{1}{m}\)
2 \(\frac{1}{{\sqrt m }}\)
3 \(\frac{1}{{{m^2}}}\)
4 \(m\)
Explanation:
At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. \( \Rightarrow \frac{1}{2}m{v^2} = \frac{{KQq}}{d} \Rightarrow d \propto \frac{1}{m}\)
NEET - 2016
PHXII12:ATOMS
356352
Statement A : The trajectory traced by an incident particle depends on the impact parameter of collision. Statement B : The impact parameter is the perpendicular distance of the initial velocity vector of the incident particle from the centre of the target nucleus.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
356348
Assertion : Large angle of scattering of alpha particles led to the discovery of atomic nucleus. Reason : Entire positive charge of atom is concentrated in the central core.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\alpha - \) particle is positively charged, so is the nucleus, so the large angle of scattering of \(\alpha - \) particle shows that the nucleus is positively charged and concentrated in the central core. So option (1) is correct.
PHXII12:ATOMS
356350
The trajectory of an \(\alpha\)-particle can be computed by employing:
1 Newton's second law of motion.
2 Coulomb's law for electrostatic force of repulsion between the \(\alpha\)-particle and the positively charges nucleus.
3 Both of these.
4 None of these.
Explanation:
For situation at sub atomic level also basic laws are applicable.
PHXII12:ATOMS
356351
When an \(\alpha - \) particle of mass \('m'\) moving with velocity \('v'\) bombards on a heavy nucleus of charge \('Ze'\), its distance of closest approach from the nucleus depends on \(m\) as:
1 \(\frac{1}{m}\)
2 \(\frac{1}{{\sqrt m }}\)
3 \(\frac{1}{{{m^2}}}\)
4 \(m\)
Explanation:
At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. \( \Rightarrow \frac{1}{2}m{v^2} = \frac{{KQq}}{d} \Rightarrow d \propto \frac{1}{m}\)
NEET - 2016
PHXII12:ATOMS
356352
Statement A : The trajectory traced by an incident particle depends on the impact parameter of collision. Statement B : The impact parameter is the perpendicular distance of the initial velocity vector of the incident particle from the centre of the target nucleus.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
356348
Assertion : Large angle of scattering of alpha particles led to the discovery of atomic nucleus. Reason : Entire positive charge of atom is concentrated in the central core.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\alpha - \) particle is positively charged, so is the nucleus, so the large angle of scattering of \(\alpha - \) particle shows that the nucleus is positively charged and concentrated in the central core. So option (1) is correct.
PHXII12:ATOMS
356350
The trajectory of an \(\alpha\)-particle can be computed by employing:
1 Newton's second law of motion.
2 Coulomb's law for electrostatic force of repulsion between the \(\alpha\)-particle and the positively charges nucleus.
3 Both of these.
4 None of these.
Explanation:
For situation at sub atomic level also basic laws are applicable.
PHXII12:ATOMS
356351
When an \(\alpha - \) particle of mass \('m'\) moving with velocity \('v'\) bombards on a heavy nucleus of charge \('Ze'\), its distance of closest approach from the nucleus depends on \(m\) as:
1 \(\frac{1}{m}\)
2 \(\frac{1}{{\sqrt m }}\)
3 \(\frac{1}{{{m^2}}}\)
4 \(m\)
Explanation:
At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. \( \Rightarrow \frac{1}{2}m{v^2} = \frac{{KQq}}{d} \Rightarrow d \propto \frac{1}{m}\)
NEET - 2016
PHXII12:ATOMS
356352
Statement A : The trajectory traced by an incident particle depends on the impact parameter of collision. Statement B : The impact parameter is the perpendicular distance of the initial velocity vector of the incident particle from the centre of the target nucleus.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
356348
Assertion : Large angle of scattering of alpha particles led to the discovery of atomic nucleus. Reason : Entire positive charge of atom is concentrated in the central core.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\alpha - \) particle is positively charged, so is the nucleus, so the large angle of scattering of \(\alpha - \) particle shows that the nucleus is positively charged and concentrated in the central core. So option (1) is correct.
PHXII12:ATOMS
356350
The trajectory of an \(\alpha\)-particle can be computed by employing:
1 Newton's second law of motion.
2 Coulomb's law for electrostatic force of repulsion between the \(\alpha\)-particle and the positively charges nucleus.
3 Both of these.
4 None of these.
Explanation:
For situation at sub atomic level also basic laws are applicable.
PHXII12:ATOMS
356351
When an \(\alpha - \) particle of mass \('m'\) moving with velocity \('v'\) bombards on a heavy nucleus of charge \('Ze'\), its distance of closest approach from the nucleus depends on \(m\) as:
1 \(\frac{1}{m}\)
2 \(\frac{1}{{\sqrt m }}\)
3 \(\frac{1}{{{m^2}}}\)
4 \(m\)
Explanation:
At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. \( \Rightarrow \frac{1}{2}m{v^2} = \frac{{KQq}}{d} \Rightarrow d \propto \frac{1}{m}\)
NEET - 2016
PHXII12:ATOMS
356352
Statement A : The trajectory traced by an incident particle depends on the impact parameter of collision. Statement B : The impact parameter is the perpendicular distance of the initial velocity vector of the incident particle from the centre of the target nucleus.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
