The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy. 

Based upon the above information, arrange the following orbitals in the increasing order of energy.

$4\mathrm{s}, 3\mathrm{s}, 3\mathrm{p}, 4\mathrm{d}$

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy. 

Based upon the above information, arrange the following orbitals in the increasing order of energy. 

$5\mathrm{p}, 4\mathrm{d}, 5\mathrm{d}, 4\mathrm{f}, 6\mathrm{s}$

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy.

Based upon the above information, arrange the following orbitals in the increasing order of energy. 

$5\mathrm{f}, 6\mathrm{d}, 7\mathrm{s}, 7\mathrm{p}$ 

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy.

Based upon the above information, solve the question given below:

Which of the following orbitals has the lowest energy?  

$4\mathrm{d}, 4\mathrm{f}, 5\mathrm{s}, 5\mathrm{p}$

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of  $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy.

Based upon the above information, solve the question given below:

Which of the following orbitals has the highest energy?  

$5\mathrm{p}, 5\mathrm{d}, 5\mathrm{f}, 6\mathrm{s}, 6\mathrm{p}$

Which of the following will not show deflection from the path on passing through an electric field?

Proton, cathode rays, electron, neutron.

Wavelengths of different radiations are given below:

$\mathrm{\lambda }\left(\mathrm{A}\right)=300\mathrm{nm} \mathrm{\lambda }\left(\mathrm{B}\right)=300\mathrm{\mu m} \mathrm{\lambda }\left(\mathrm{C}\right)=3\mathrm{nm} \mathrm{\lambda }\left(\mathrm{D}\right)=30\mathrm{A}°$

Arrange these radiations in the increasing order of their energies.