Given X-ray spectrum is for a coolidge tube having accelerating potential `V`. If accelerating potential is decreased to `V//4` , then `Deltalambda=lambda-lambda_(c)` becomes four times with change in anode element . If `Z` is the atomic number of the original element, them the atomoc number of new element is (neglect screening effect)

Options

A. `Z`
B. `Z//2`
C. `2Z`
D. `Z//3`

Answer

B

Explanation

In the first case : `Deltalambda =4/(3RZ^(2))(hc)/(eV)(therefore1/lambda_(ka)=RZ^(2)(1/1-1/4))`
in the second case`4Delta=4/(3RZ_(2)^(2))-(4hc)/(eV)RightarrowZ_(2)=Z/2`