To excite the spectral line of wavelength 4960 Å of an atom, an excitation energy of 7.7 eV is required. The ground state energy of the atom is 10.5 eV.
The energies of two levels involved in the emission of 4960 Å line are :( Assume hc = 1240 eV -nm , where h is planck's constant and c is speed of light . )

Options

A. 14.2 eV, 16.1 eV
B. 12.2 eV, 18.2 eV
C. 15.7 eV,20.5 eV
D. 15.7 eV, 18.2 eV

Answer

D

Explanation

Given , ground state energy of an atom,
`E_g = 10.5 eV`
excitation energy = 7.7 eV and wavelengths of spectral line `lambda ` = Å
As , it is given that excitation of 7.7 eV required to limit electromagnetic wave (EMW) of wavelength 4960 Å, then the energy of excited state is
`E_1 =E_g +7.7eV = 10.5 +7.7`
`E_1 = 18.2 eV`
Now, the energy of photon of wavelength 4960 Å is
`E=(hc)/(lambda)= (12400)/(4960) = 2.5 eV`
So , the next possible spectral line emitting state energy ,
`E_2 =E_1 -2.5 eV= 18.2 - 2.5 = 15.7 ev`
So, the energy of two states is 15.7 eV and 18.2 eV