You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Molecular Speeds - Vrms, Vavg, Vmps

Find the freq...

Question

Find the frequency of revolution of the electron in the second orbit of H-atom.

802 \times 10^{14} / s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

606 \times 10^{10} / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

226.6 \times 10^{- 10} / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

116.6 \times 10^{15} / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $802 \times 10^{14} / s$

Suggest Corrections

Similar questions

Q. If the frequency or revolution of the electron in first Bohr orbit of a H-atom is

6.6 \times 10^{15}

Hz, the frequency in the next orbit is

Q. The time period of revolution of electron in its ground state orbit in a hydrogen atom is

1.6 \times 10^{- 16} s .

The frequency of revolution of the electron in its first excited state (in

s^{- 1}

) is :

Q. In a hydrogen-like atom, an electron is orbiting in an orbit having quantum number

n

. Its frequency of revolution is found to be

13.2 \times 10^{15} H z

. Energy required to move this electron from the atom to the above orbit is

54.4 e V

. In a time of

7

nano second the electron jumps back to orbit having quantum number

n / 2

. If

τ

be the average torque acted on the electron during the above process, then

τ = (10 + x) \times 10^{27}

N m

. Find the value of

x

Given:

h / λ = 2.1 \times 10^{- 34} J - s

, frequency of revolution of electron in the ground state of

H

atom

v_{0} = 6.6 \times 10^{15}

and ionization energy of

H

atom

E_{0} = 13.6 e V

Q. In a hydrogen like atom, an electron is revolving in an orbit having quantum number

n

. Its frequency of revolution is found to be

1.32 \times 10^{15} Hz .

Energy required to completely remove this electron to infinity from the above orbit of atom is

54.4 eV

. In a time interval of

7

nanoseconds the electron jumps back to orbits having quantum number

n / 2.

If the average torque acting on the electron during the above process is

τ Nm,

then find the value of

τ \times 10^{27}

(given

\frac{h}{π} = 2.1 \times 10^{- 34} J-s,

) frequency of revolution of electron in the ground state of H-atom

= 6.6 \times 10^{15} Hz

_______

Q. If the velocity of electron in the first Bhors orbit of H atoms is s

2.19 \times 10^{6} m / s

.calculate the velocity of electron in Bohrs second orbit?

Join BYJU'S Learning Program

Find the frequency of revolution of the electron in the second orbit of H-atom.

The correct option is A 802×1014/s

The correct option is A $802 \times 10^{14} / s$