A hot gas emites radition of wavelength `46.0nm ,82.8nm and 103.5nm` only 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

A, B

Explanation

Energy in ground state is the energy acquired in the transition of 2^(nd) excitation state to ground state. As 2^(nd) excitation state is taken as zero level.
`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`