Chemistry Class 11 NCERT Solutions: Chapter 2 Structure of Atom Part 18

Q: 56 Calculate the wavelength for the emission transition if it starts from the orbit having radius Equation nm and ends at Equation pm. Name the series to which this transition belongs and the region of the spectrum.

Answer:

The radius of the Equation orbit of hydrogen-like particles is given by,

Equation

For radius Equation

Equation

Equation

Equation

Equation

Equation

Similarly,

Equation

Equation

Equation

Equation

Equation

Equation

Thus, the transition is from the Equation orbit to the Equation orbit. It belongs to the Balmer series.

Wave number Equation for the transition is given by,

Equation

Equation

Equation

Equation Wavelength Equation associated with the emission transition is given by,

Equation

Equation

Equation

Equation

Q: 57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is Equation calculate de Broglie wavelength associated with this electron.

Answer:

From de Broglie’s equation,

Equation

Equation

Equation

Equation

Equation

Q: 58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used hare is 800 pm, calculate the characteristic velocity associated with the neutron.

Answer:

From de Broglie’s equation,

Equation

Equation

Where,

V= velocity of particle (neutron)

h =Planck’s constant

m = Mass of particle (neutron)

Equation

Substituting the values in the expression of velocity (v),

Equation

Equation

Equation

Equation

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