# 1) if the energy radius and velocity of an electron in the ground state of hydrogen atom are: E=-13.6ev, R 0.53A and V 2.2*10®6M/s respectively, calculate the energy radius and velocity for an electron of energy.n=3,n=5,n=10.2) If the excitation energy for two given levels of an atom is 10.2ev,calculate the wavelength of the emitted photon and the frequency. 3) If the energy level for n=1 is -13.6ev,find the energy level for n=3 and n=8 obtained. The excitation energy between the two levels in joules. 4) An atom excited to an energy level E2 falls to a ground state, calculate the frequency and the wavelength of the emitted photon. (h=6.63*10®-34js)E2=-1.51evE1=-13.6ev.

Vikas TU
14149 Points
7 years ago
Dear Student,
As from the given data,
E=-13.6eV
R=0.53A
V=2.2x106
En=E/n2
E3=-1.511 eV
E5=-0.544eV
E10=-0.136eV

Rn=Rxn2
R3=0.53x9 =4.77A
R5=0.53x25=13.25A
R10=53A

Vn=V/n
V3=2.2x106/3=0.733x106m/s
V5=2.2x106/5=0.44x106m/s
V10=0.22x106 m/s.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)