If 20 electrons are present in $4^{\mathit{th }}$ orbit, then the atomic number of that element is 

If there are 20 electrons present in $4^{\mathit{th }}$ orbit, then the $4^{\mathit{th }}$ orbit has electrons filled as ${4s}^2{4p}^{6 }{4d}^{10}{4f^2}^{ }$. According to Aufbau principle and Hund's rule the element would belong to the inner transition series as the last electron enters into f-subshell and would have electronic configuration as ${1s}^2{2s}^{2 }{2p}^6{3s^2}^{ }{3p}^6{4s}^{2 }{3d}^{10}{4p^6}^{ }{5s}^2{4d}^{10 }{5p}^6{6s^2{4f}^2}^{ }$.
The atomic number of the element is 58 and the name of the element is Cerium (Ce).
 If 20 electrons are present in $4^{\mathit{th }}$ orbit, then the atomic number of that element is 58.