Question

A hot gas emits radiation of wavelengths 46.0 nm, 82.8 nm and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.

Answer 1

Energy in ground state is the energy acquired in the transition of second excited state to ground state.

As second excited state is taken as zero level.

$E=\frac{hc}{{\lambda }_1}$

$=\frac{4.14\ast {10}^{-15}\ast 3\ast {10}^8}{46\ast {10}^{-9}}=\frac{1242}{46}=27eV$

Again energy in the first excited state

$E=\frac{hc}{{\lambda }_{11}}$

$=\frac{4.14\ast {10}^{-15}\ast 3\ast {10}^8}{103.5}=12eV$