Question

The difference in angular momentums associated with an electronic transition in hydrogen atom is equal to $3h/2\pi$. The atom contains only K, L, M and N shells. Calculate the energies and radii associated with the orbits.

• A. $-13.6$ eV and $0.314$ Angstrom, $0.08$ eV and 8.48 Angstrom
• B. $-13.6$ eV and $0.529$ Angstrom, $-0.08$ eV and $8.48$ Angstrom
• C. $13.6$ eV and $0.529$ Angstrom, $0.08$ eV and $8.48$ Angstrom
• D. $27.2$ eV and $0.529$ Angstrom, $0.08$ eV and $8.48$ Angstrom

Answer 1

The correct option is B $-13.6$ eV and $0.529$ Angstrom, $-0.08$ eV and $8.48$ Angstrom

Angular momentum is given as: $\frac{nh}{2\pi }$
Therefore angular momentum of electron in K shell = $\frac{h}{2\pi }$
Angular momentum of electron in L shell = $\frac{2h}{2\pi }$
Angular momentum of electron in M shell = $\frac{3h}{2\pi }$
Angular momentum of electron in N shell = $\frac{4h}{2\pi }$
Thus the electronic transition is associated with K and N shell.
Energy of electron in K shell $(n=1)=$ $-13.6\frac{Z^2}{n^2}$ $=-13.6eV$
Energy of electron in N shell $(n=4)=$ $-13.6\left(\frac{1^2}{4^2}\right)$$=-0.08eV$
Radius of orbit in K shell = $0.529\times 1^2$ $=0.529$ $A^{\circ }$
Radius of orbit in N shell = $0.529\times 4^2$ $=8.48$ $A^{\circ }$