Question

Moseley equation is represented as √v=a(z-b) where a and b are constant if oa = 1 then atomic number of the element showing frwquency of 400 hz is

Answer 1

Answer:

The answer is Z = 21

Step-by-step explanation:

Mosely's Equation is √v = a(Z-b)  

Here v is the frequency, a and b are constants.

Now we are given that a=b=1 & v=400 Hz so substituting th values in the above equation we get,  

√400 = 1(Z-1)  

20 = Z-1  

Therefore Z = 21

Z =21 is the atomic number of Sc which is Scandium.

Answer 2

The answer is Z = 21

Step-by-step explanation:

Mosely's Equation is √v = a(Z-b)  

Here v is the frequency, a and b are constants.

Now we are given that a=b=1 & v=400 Hz so substituting th values in the above equation we get,  

√400 = 1(Z-1)  

20 = Z-1  

Therefore Z = 21

Z =21 is the atomic number of Sc which is Scandium.