Home
Class 11
CHEMISTRY
A particle moving with a velocity of 6.6...

A particle moving with a velocity of `6.626 xx 10^(7)` m/s has a de Broglie wavelength of `1 "Å"` in a circular path of radius `0.529 "Å"` . The angular momentum of particle is (h = `6.626 xx 10^(-34) J xx "sec"`):

Promotional Banner

Similar Questions

Explore conceptually related problems

The de Broglie wavelength associated with a moving particle changes from 0.2Å to 0.4Å . Determine the change in momentum of the particle. (h = 6.63xx10^(-34) J.s) .

Energy of an electron in an excited hydrogen atom is -3.4eV . Its angualr momentum will be: h = 6.626 xx 10^(-34) J-s .

Energy of an electron in an excited hydrogen atom is -3.4eV . Its angualr momentum will be: h = 6.626 xx 10^(-34) J-s .

The de-Broglie wavelength of an electron is 600 nm . The velocity of the electron is: (h = 6.6 xx 10^(-34) J "sec", m = 9.0 xx 10^(-31) kg)

The wavelength of de - Broglie wave is 2 mu m , then its momentum is ( h = 6.63 xx 10^(-34 J-s)

The wavelength of de - Broglie wave is 2 mu m , then its momentum is ( h = 6.63 xx 10^(-34 ) J-s)

A body of mass x g is moving with a velocity of 100m/s. It de Broglie wavelength is 6.62xx10^(-35)m . Hence x is (h=6.62xx10^(-34)J-s)

A particle having a de Broglie wavelength of 1.0 A^(0) is associated with a momentum of (given h=6.6xx10^(-34)Js )

The wavelength of a moving body of mass 0.1 mg is 3.31 xx 10^(-29) m. Calculate its kinetic energy (h = 6.626 xx 10^(-34) Js) .

The wavelength of a moving body of mass 0.1 mg is 3.31 xx 10^(-29) m. Calculate its kinetic energy (h = 6.626 xx 10^(-34) Js) .