A hypothetical atom has four defined energy levels as given in the table below: Principal quantum number ( n ) Energy ( E ) n=1 -12 eV n=2 -6 eV n=3 -2 eV n=4 -1 eV Which of the following transitions will result in the emission of a photon of wavelength 310 nm ? (Take hc = 1240 eV nm )

Step-by-step solution

The energy of the emitted photon is given by: E = hc Substituting the given values: E = 1240 eV nm 310 nm = 4 eV Now, we check the energy differences between the given levels for emission (transition from higher n to lower n ): For n = 4 to n = 1 : E = -1 eV - (-12 eV ) = 11 eV For n = 4 to n = 2 : E = -1 eV - (-6 eV ) = 5 eV For n = 3 to n = 2 : E = -2 eV - (-6 eV ) = 4 eV For n = 2 to n = 1 : E = -6 eV - (-12 eV ) = 6 eV The transition that yields an energy difference of 4 eV is from n = 3 to n = 2 . Answer: n =