When a certain element is bombarded with high-energy electrons, kα x-rays that have an energy of 2611 ev are emitted. determine the atomic number z of the element. use the bohr model as necessary?

Use Moseley's Law for Kalpha. 

nu = (2.47 x 10^15) (Z - 1)^2 

We know that E = h*nu, so... 
nu = E / h 

substitute, rearrange, etc, you get... 

Z = sqrt( (E/h)/(2.47*10^15) ) + 1 

Answer yields 18.98 ... ~ 19, hence Potassium, like I wrote earlier in reference to NIST tables.