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The de brogile wavelength associated with electrons revolving round the nucleus in a hydrogen atomin a ground state will.(1)0.3(2)3.3(3)6.62(4)10?
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The de brogile wavelength associated with electrons revolving round th...
The de Broglie wavelength associated with electrons revolving around the nucleus in a hydrogen atom in the ground state can be calculated using the de Broglie wavelength equation:

λ = h / p

Where:
λ is the de Broglie wavelength
h is the Planck's constant (6.626 × 10^-34 Js)
p is the momentum of the electron

To determine the momentum of the electron, we need to use the Bohr model of the hydrogen atom. According to the Bohr model, the momentum of the electron is given by:

p = mv

Where:
m is the mass of the electron (9.11 × 10^-31 kg)
v is the velocity of the electron

To find the velocity of the electron, we can use the equation for the centripetal force in circular motion:

Fc = mv^2 / r

Where:
Fc is the centripetal force
m is the mass of the electron
v is the velocity of the electron
r is the radius of the electron's orbit

In the ground state of the hydrogen atom, the electron revolves around the nucleus in the first energy level, which corresponds to the Bohr radius (a0) of approximately 0.529 × 10^-10 m.

The centripetal force in this case is provided by the electrostatic attraction between the electron and the nucleus, given by Coulomb's law:

Fc = ke^2 / r^2

Where:
ke is the electrostatic constant (8.988 × 10^9 Nm^2/C^2)
e is the elementary charge (1.602 × 10^-19 C)

By equating these two expressions for the centripetal force, we can solve for the velocity of the electron:

mv^2 / r = ke^2 / r^2

v^2 = ke^2 / mr

v = √(ke^2 / mr)

Now we can substitute the values into the equation for the de Broglie wavelength:

λ = h / (mv)

Substituting the values for h, m, and v, we can calculate the de Broglie wavelength associated with an electron in the ground state of a hydrogen atom.

The correct answer for the de Broglie wavelength is (3) 6.62.
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Read the following text and answer the following questions on the basis of the same:Electron Microscope Electron microscopes use electrons to illuminate a sample. In Transmission Electron Microscopy (TEM), electrons pass through the sample and illuminate film or a digital camera.Resolution in microscopy is limited to about half of the wavelength of the illumination source used to image the sample. Using visible light the best resolution that can be achieved by microscopes is about ~200 nm. Louis de Broglie showed that every particle or matter propagates like a wave. The wavelength of propagating electrons at a given accelerating voltage can be determined byThus, the wavelength of electrons is calculated to be 3.88 pm when the microscope is operated at 100 keV, 2. 74 pm at 200 keV and 2.24 pm at 300 keV. However, because the velocities of electrons in an electron microscope reach about 70% the speed of light with an accelerating voltage of 200 keV, there are relativistic effects on these electrons. Due to this effect, the wavelength at 100 keV, 200 keV and 300 keV in electron microscopes is 3.70 pm, 2.51 pm and 1.96 pm, respectively.Anyhow, the wavelength of electrons is much smaller than that of photons (2.5 pm at 200 keV). Thus if electron wave is used to illuminate the sample, the resolution of an electron microscope theoretically becomes unlimited. Practically, the resolution is limited to ~0.1 nm due to the objective lens system in electron microscopes. Thus, electron microscopy can resolve subcellular structures that could not be visualized using standard fluorescence microscopy.Q. Wavelength of electron as wave at accelerating voltage 200 keV is

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The de brogile wavelength associated with electrons revolving round the nucleus in a hydrogen atomin a ground state will.(1)0.3(2)3.3(3)6.62(4)10?
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