Saved Bookmarks
| 1. |
A hot gas emites radition of wavelength 46.0nm ,82.8nm and 103.5nmonly Assume that the atoms have only two excited state and the difference between consecutive energy levels decrease as energy is increased Taking the energy of the higest energy state to be zero find the energies of the ground state and the first excited state |
|
Answer» `E= (hc)/(lambda_1)=(4.14 XX 10^-15 xx 10^8)/(46 xx 10^-9)` `=1242/46 = 27eV` Again energy in the first excitation state `E=(hc)/(lambda_n)` `=(4.14 xx 10^-15 xx 10^8)/(103.5)` `=12.eV` |
|