In a hydrogen like atom, an electron is revolving in an orbit having quantum number n. Its frequency of revolution is found to be 1.32×1015 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×1027 (given hπ=2.1×10−34 J-s,) frequency of revolution of electron in the ground state of H-atom =6.6×1015 Hz_______

In a hydrogen like atom, an electron is revolving in an orbit having q

1.	In a hydrogen like atom, an electron is revolving in an orbit having quantum number n. Its frequency of revolution is found to be 1.32×1015 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×1027 (given hπ=2.1×10−34 J-s,) frequency of revolution of electron in the ground state of H-atom =6.6×1015 Hz_______
Answer» 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 $= 6.6 \times 10^{15} Hz$ _______