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. - Chemistry


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 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron.


Solution 1

From de Broglie’s equation,

`lambda = "h"/("mv")`

`lambda = (6.626 xx 10^(-34) "Js")/(9.10939 xx 10^(-31)(1.6xx10^6 " ms"^(-1)))`

= 4.55 × 10–10 m

λ = 455 pm

∴ de Broglie’s wavelength associated with the electron is 455 pm.

Solution 2

λ = h/mv = 6.626×10-34 kgm2s-1/(9.11×10-31 kg) (1.6×10ms-1) = 4.55×10-10 m = 455 pm

  Is there an error in this question or solution?
Chapter 2: Structure of Atom - EXERCISES [Page 72]


NCERT Chemistry Part 1 and 2 Class 11
Chapter 2 Structure of Atom
EXERCISES | Q 2.57 | Page 72


Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms-1.

The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.

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

The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature.

Assertion (A): All isotopes of a given element show the same type of chemical behaviour.

Reason (R): The chemical properties of an atom are controlled by the number of electrons in the atom.

What will be the de Broglie wavelength (in metres) of an electron moving with a velocity of 1.2 × 105 m s–1?

A proton and a Li3+ nucleus are accelerated by the same potential. If λLi and λP denote the de Broglie wavelengths of Li3+ and proton respectively, then the value of `λ_("Li")/(λ_"P")` is x × 10−1. The value of x is ______. [Rounded-off to the nearest integer]

[Mass of Li3+ = 8.3 mass of proton]

When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is ______ A°. (Round off to the Nearest Integer)

[Use: `sqrt3` = 1.73, h = 6.63 × 10−34 Js, me = 9.1 × 10−31 kg, c = 3.0 × 108 ms−1, 1eV = 1.6 × 10−19 J]

The de Broglie's wavelength of electron emitted by metal will be ______ Å whose threshold frequency is 2.25 × 1014 Hz when exposed to visible radiation of wavelength 500 nm.

The de Broglie wavelength of an electron in the 4th Bohr orbit is ______.


      Forgot password?
Use app×