Using the Bohr model, what is the ratio of the energy of the nth orbit of a triply ionized beryllium atom (Be3+, Z = 4) to the energy of the nth orbit of a hydrogen atom (H)? Enth orbit, Be / Enth orbit, H

Respuesta :

Answer:

16

Explanation:

solution:

by taking the ratio of the energy E_n,be of the nth orbit of a beryllium atom

Z_be=4 to the energy E_n,h of the nth orbit of a hydrogen (Z_h=1) atom  gives

E_n,B/E_n,H=-(2.18*10^-18)*Z^2_BE/-(2.18*10^-18)*Zh^2/n^2

                    =Z^2_BE+/Z^2_H

                    =(4)^2/(1)^2

                    = 16