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Post a LessonAnswered on 10 Apr Learn Structure of Atom
Sadika
Rutherford's α-particle scattering experiment provided significant insights into the structure of the atom. Let's analyze each option:
(a) Most of the space in the atom is empty. - This conclusion could be derived from Rutherford's experiment. The fact that most α-particles passed through the atom undeflected indicated that the majority of the atom is empty space.
(b) The radius of the atom is about 10−1010−10 m while that of the nucleus is 10−1510−15. - This conclusion could be derived from Rutherford's experiment. The scattering pattern of α-particles suggested that the nucleus, where most of the atom's mass is concentrated, is very small compared to the overall size of the atom.
(c) Electrons move in a circular path of fixed energy called orbits. - This conclusion could not be directly derived from Rutherford's experiment. Rutherford's experiment focused on the nucleus of the atom and did not provide direct evidence for the existence of electron orbits. This concept was later proposed by Niels Bohr based on the quantization of angular momentum in the hydrogen atom.
(d) Electrons and the nucleus are held together by electrostatic forces of attraction. - This conclusion could be derived from Rutherford's experiment. The fact that some α-particles were deflected suggested the presence of a positively charged nucleus that interacted with the negatively charged electrons, leading to the conclusion that electrons and the nucleus are held together by electrostatic forces of attraction.
Therefore, the conclusion that could not be directly derived from Rutherford's α-particle scattering experiment is:
(c) Electrons move in a circular path of fixed energy called orbits.
Answered on 10 Apr Learn Structure of Atom
Sadika
To calculate the wavelength of the cricket ball, we'll use the de Broglie wavelength formula:
λ=hpλ=ph
Where:
The momentum (pp) of the cricket ball can be calculated using the formula:
p=mvp=mv
Where:
Given:
First, we need to convert the velocity to meters per second (m/s):
100 km/h=100×1000/3600 m/s
=100000/3600 m/s
=27.78 m/s
Now, we can calculate the momentum (pp):
p=(0.1 kg)×(27.78 m/s)p=(0.1kg)×(27.78m/s) p=2.778 kg m/sp
Now, we can calculate the wavelength (λλ):
λ=6.626×10−34 m2kg/s2.778 kg m/sλ=2.778kg m/s6.626×10−34m2kg/s λ≈2.384×10−34 mλ≈2.384×10−34m
Explanation: The wavelength of the cricket ball is incredibly small (on the order of 10−34meters), which is far too small to be observable or relevant at the macroscopic scale of everyday objects. Therefore, despite having a wave-like nature according to de Broglie's hypothesis, it does not exhibit wave-like behavior in its motion because the wavelength is negligible compared to the size of the cricket ball and the distances involved in its motion.
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Sadika
The wave number (ν~ν~) associated with the transition in the Balmer series when the electron moves to the n=4n=4 orbit can be calculated using the Rydberg formula:
ν~=RH(1n12−1n22)ν~=RH(n121−n221)
Where:
Plugging in the values:
ν~=109677 (1/2sqr −1/4sqr)
ν~=109677 (1/4−1/16)
ν~=109677 (0.25−0.0625)
ν~=109677×0.1875
ν~=20573 .625 cm−1
ν~=20573.625cm−1
Answered on 01/01/2022 Learn Structure of Atom
Mallela Harish
Lecturer in Chemistry with 6+ years of teaching experience
Hello Kasireddy,
The electronic confugaration of Copper (Cu) is [Ar] but not
.
This is because of
stability of completely filled d sub-shell.
The completely filled & hlaf-filled sub-shells are stable.
The causes of stability of completely filled & half filled sub-shells are due to following reasons
1) symmetrical distribution of electrons
2) exchange energy
due to above reasons Copper has [Ar] electronic confuguratiion.
Thank you
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Answered on 10 Apr Learn Structure of Atom
Sadika
Out of the options provided, the neutron will not show deflection from the path on passing through an electric field.
Neutrons are electrically neutral particles, meaning they have no electric charge. Therefore, they are not affected by electric fields and do not experience any deflection when passing through them.
On the other hand, protons, electrons, and cathode rays (which are electrons) are all charged particles and will experience deflection when passing through an electric field. Protons, being positively charged, would deflect in the opposite direction of electrons, which are negatively charged.
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Sadika
To calculate the total number of angular nodes and radial nodes present in a 3p orbital, we first need to understand the concept of angular nodes and radial nodes.
Angular Nodes: These nodes arise due to the angular part of the wave function (spherical harmonics) and are related to the angular momentum quantum number, l. For p orbitals, l = 1, which means there is 1 angular node.
Radial Nodes: These nodes arise due to the radial part of the wave function (radial probability distribution function) and are related to the principal quantum number, n, and the azimuthal quantum number, l. The formula for radial nodes is given by n - l - 1. For a 3p orbital, n = 3 and l = 1, so the number of radial nodes is 3 - 1 - 1 = 1.
Now, we can calculate the total number of nodes by adding the angular nodes and radial nodes:
Total number of nodes = Angular nodes + Radial nodes Total number of nodes = 1 (angular nodes) + 1 (radial nodes) = 2 nodes
Therefore, a 3p orbital has a total of 2 nodes, with 1 angular node and 1 radial node.
Answered on 01/01/2022 Learn Structure of Atom
Mallela Harish
Lecturer in Chemistry with 6+ years of teaching experience
Answered on 10 Apr Learn Structure of Atom
Sadika
The wavelength of a matter wave is inversely proportional to the momentum of the particle, according to the de Broglie equation:
λ=hpλ=ph
Where:
Given that the particles are traveling at the same speed, their momentum will be directly proportional to their mass.
The mass of the particles in order of increasing mass:
Since all particles are traveling at the same speed, the particle with the smallest mass will have the shortest wavelength.
So, the electron (option a) will have the shortest wavelength among the given options.
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Mallela Harish
Lecturer in Chemistry with 6+ years of teaching experience
Answer is A
Orbital angular momentum depends on the value of azimuthal quantum number.
Azimuthal Quantum Number describes the shape of the orbital.
It is represented as 'l'.
The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
The orbital angular momentum of an electron is given by
l = azimuthal quantum number
h = plank's constant
Thus Orbital angular momentum depends on the value of azimuthal quantum number i.e the value of l.
Thank you.
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