The X-ray wavelength of `L_(a)`line of platinum (Z=78) is 1:30 Å.The X-ray wavelength of `L_(a)`line of Molybdeum (Z=42) is

Options

A. 5.41Å
B. 4.59Å
C. 2.70Å
D. 1.35Å

Answer

B

Explanation

Atomic number of Platinum ,z=78
Atomic number of Molybdenum ,z= 42
The expression for the X- ray wavelength of a particular line is given as
`(1)/(lamda)=R(Z-1)^(z)`
`therefore (lamda_(1))/lamda_(2)(Z_(2)-(1)^(2))/(Z_(1)-1)^(2)`
`(1.30)/(lamda^(2))=(42-1)^(2)/((78-1)^(2))` (becauseX-ray wavelength
of platinum =1.30)
`implies (1.30)/(lamda^(2)0=(1681)/(5929)`
`implies 1.30=0.283 `implieslamda_(2)=(1.30)/(0.283)therefore lamda_(2)=4.59A^(o)`