ExamBro
ExamBro
TS EAMCET · Physics · Nuclear Physics

To excite the spectral line of wavelength 4960 A 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 A line are (Assume he =1240 eV nm, where h is Planck's constant and c is speed of light)

  1. A 14.2eV,16.1eV
  2. B 12.2eV,18.2eV
  3. C 15.7eV,20.5eV
  4. D 15.7eV,  18.2eV
Verified Solution

Answer & Solution

Correct Answer

(D) 15.7eV,  18.2eV

Step-by-step Solution

Detailed explanation

Energy of excited state = Ground state energy + Excitation energy E1=10.5+7.7=18.2 eV Energy of photon is, E=hcλ=124004960=2.5 eV ∴E2=E1-2.5 eV=18.2-2.5=15.7 eV