The Ka line from molybdenum (atomic number =42 ) has a wavelength of 0.7078 mathringA. The wavelength of Ka line of zinc (atomic number =30 ) will be

Question

The Ka line from molybdenum (atomic number =42 ) has a wavelength of 0.7078 mathringA. The wavelength of Ka line of zinc (atomic number =30 ) will be (A) 1 mathringA (B) 1.3872 mathringA (C) 0.3541 mathringA (D) 0.5 mathringA

Tardigrade Admin · Accepted Answer

Wavelength λ=0.7078 mathringA EnergyE=(h c/λ) (z2/n2)=(h c/λ) ∴ (z12/z22)=(λ2/λ1) ((42)2/(30)2)=(λ2/0.7078) λ=1.3873 mathringA

Q. The $K_{a}$ line from molybdenum (atomic number $= 42$ ) has a wavelength of $0.7078 \overset{˚}{A}$ . The wavelength of $K_{a}$ line of zinc (atomic number $= 30$ ) will be

$1 \overset{˚}{A}$

$1.3872 \overset{˚}{A}$

$0.3541 \overset{˚}{A}$

$0.5 \overset{˚}{A}$

Wavelength $λ = 0.7078 \overset{˚}{A}$
Energy $E = \frac{h c}{λ}$
$\frac{z ^{2}}{n ^{2}} = \frac{h c}{λ}$
$∴ \frac{z _{1}^{2}}{z _{2}^{2}} = \frac{λ _{2}}{λ _{1}}$
$\frac{( 42 ) ^{2}}{( 30 ) ^{2}} = \frac{λ _{2}}{0.7078}$
$λ = 1.3873 \overset{˚}{A}$

Q. The Ka​ line from molybdenum (atomic number =42 ) has a wavelength of 0.7078A˚. The wavelength of Ka​ line of zinc (atomic number =30 ) will be

1A˚

1.3872A˚

0.3541A˚

0.5A˚