The Radius of an Hydrogen atom is `2.116A^(0)` .The velocity of an electron is `x times10^(y)m/s` ,then y-x is

The radius of the fourth orbit in hydrogen atom is 0.85 nm. Calculate the velocity of the electron in this orbit.

The circumference of second orbit of hydrogen atom, if the wavelength of electron is 5 xx 10^(-9)m will be

The radius of the second orbit of an electron in hydrogen atom is 2.116A . The de Broglie wavelength associated with this electron in this orbit would be

The radius of hydrogen atom in its ground state is 5.3 xx 10^(-11)m . After collision with an electron it is found to have a radius of 21.2 xx 10^(-11)m . What is the principle quantum number of n of the final state of the atom ?

The radius of hydrogen atom in its ground state is 5.3 xx 10^(-11)m . After collision with an electron it is found to have a radius of 21.2 xx 10^(-11)m . What is the principle quantum number of n of the final state of the atom ?

The circumference of the first Bohr orbit in H atom is 3.322 xx 10^(-10) m .What is the velocity of the electron of this orbit ?

In Bohr.s model of hydrogen atom, radius of the first orbit of an electron is r_(0) . Then, radius of the third orbit is:

The velocity of an electron in the first Bohr orbit of hydrogen atom is 2.19 xx 10^(6)ms^(-1) . Its velocity in the second orbit would be

The velocity of an electron in the orbit of hydrogen atom is 5.47xx10^(5) m/sec. Total number of waves formed by the electron in one complete revolution is

If the radius of the first orbit of the hydrogen atom is 0.53 Å , then the de-Broglie wavelength of the electron in the ground state of hydrogen atom will be