Arrange in decreasing order the energy of 2s orbital in the following atoms H, Li, Na, K

As we go down the group IA, there is increase in the number of shells so size of atom increases and energy of 2s orbital decreases.

Hence the correct order is, ${\mathrm{E}}_{2\mathrm{s}}\left(\mathrm{H}\right)>{\mathrm{E}}_{2\mathrm{s}}\left(\mathrm{Li}\right)>{\mathrm{E}}_{2\mathrm{s}}\left(\mathrm{Na}\right)>{\mathrm{E}}_{2\mathrm{s}}\left(\mathrm{K}\right)$