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Structure of Atom

Structure of Atom relates to CBSE/Class 11/Science/Chemistry

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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Structure of Atom

Sadika

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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/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... read more

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.

 
 
 
 
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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Structure of Atom

Sadika

To calculate the wavelength of the cricket ball, we'll use the de Broglie wavelength formula: λ=hpλ=ph Where: λλ is the wavelength hh is the Planck constant (6.626×10−34 m2kg/s) pp is the momentum The momentum (pp) of the cricket ball can be calculated... read more

To calculate the wavelength of the cricket ball, we'll use the de Broglie wavelength formula:

λ=hpλ=ph

Where:

  • λλ is the wavelength
  • hh is the Planck constant (6.626×10−34 m2kg/s)
  • pp is the momentum

The momentum (pp) of the cricket ball can be calculated using the formula:

p=mvp=mv

Where:

  • mm is the mass of the cricket ball (in kg)
  • vv is the velocity of the cricket ball (in m/s)

Given:

  • Mass of the cricket ball (mm) = 100 g = 0.1 kg
    Velocity of the cricket ball (vv) = 100 km/h

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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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Structure of Atom

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: RHRH is the Rydberg constant (109677 cm−¹) n1=2n1=2 is... read more

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(n121n221)

Where:

  • RHRH is the Rydberg constant (109677 cm−¹)
  • n1=2n1=2 is the initial orbit
  • n2=4n2=4 is the final orbit

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


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Answered on 01/01/2022 Learn CBSE/Class 11/Science/Chemistry/Structure of Atom

Mallela Harish

Lecturer in Chemistry with 6+ years of teaching experience

Hello Kasireddy, The electronic confugaration of Copper (Cu) is 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... read more

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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