An energy of 68.0 eV is required to excite a hydrogen-like atom in its second Bohr energy level to third energy level the charge of nucleus is Ze . The wavelength of a radiation required to eject the electron from first orbit to infinity is

Step-by-step solution

Given, n₁=2, n₂=3 and E=68 eV The difference in energies of two orbits E=13.6 Z^2 [ 1 n₁^2 - 1 n₂^2 ] where, n₁ n₂ aligned 68 & =13.6 Z^2 [ 1 2^2 - 1 3^2 ] & =13.6 Z^2 [ 1 4 - 1 9 ] & =13.6 Z^2 [ 9-4 36 ] & =13.6 Z^2 5 36 aligned aligned Z^2 & = 68 36 13.6 5 =36 Z & =6 aligned Wavelength of photon, 1 =Z^2 R [ 1 n₁^2 - 1 n₂^2 ]n₁=1 and n₂= 1 =6^2 R [ 1 1^2 - 1 ]=36 R = 1 36 R = 1 36 1.097 10^7 aligned & = 10⁻⁷ 39.5 ~m & =0.025 10⁻⁷ ~m & =2.5 ~nm aligned