The K- X- rays arising from a cobalt Z-27 target have a wavelength of 179 pm- The K- X rays arising from a nickle target Z-28 is

Λ∝1/(Z 1)^2 λ_Ni/λ_Co=(Z_Co 1/Z_Ni 1)^2=(27 1/28 1)^2 λ_Ni=(26)^2/(27)^2λ_Co nbsp; (λ_Ni lt; λ_Co) nbsp; ...

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Standard XII

Physics

Stopping Potential Revisited

The Kα X- ray...

Question

The $K_{α}$ X- rays arising from a cobalt ( $Z = 27$ ) target have a wavelength of 179 $p m$ . The $K_{α}$ X rays arising from a nickle target (Z=28) is

> 179 $p m$

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< 179 $p m$

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= 179 $p m$

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None of these

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Solution

The correct option is B
< 179 $p m$

$λ \propto \frac{1}{(Z - 1)^{2}}$

$\frac{λ_{N i}}{λ_{C o}} = (\frac{Z_{C o} - 1}{Z_{N i} - 1})^{2} = (\frac{27 - 1}{28 - 1})^{2}$
$λ_{N i} = \frac{(26)^{2}}{(27)^{2}} λ_{C o}$ $(λ_{N i} < λ_{C o})$

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Similar questions

The $K_{α}$ X- rays arising from a cobalt ( $Z = 27$ ) target have a wavelength of 179 $p m$ . The $K_{α}$ X rays arising from a nickle target (Z=28) is

X -

rays emitted from a copper target and a molybdenum target are found to contain a line of wavelength

22.85 n m

attributed to the

K_{α}

line of an impurity element. The

K_{α}

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(Z = 29)

and molybdenum

(Z = 42)

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15.42 n m

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Q. X-ray from a tube with a target A of atomic number Z shows strong

K_{α}

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for target A and

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and

λ_{2}

for two impurities respectively.

\frac{λ_{0}}{λ_{1}} = 4

and

\frac{λ_{0}}{λ_{2}} = \frac{1}{4}

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K_{α}

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Q. X-rays emitted from a copper target and a molybdenum target are found to contain a line of wavelength

22.85

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K_{α}

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= 29

) and molybdenum (Z

= 42

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15.42

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7.12

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v^{1 / 2} = a (Z - b)

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Q. The wavelength of

K_{α} X-ray

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21.3 pm

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K_{α} X-ray

(h c = 1242 eV-nm)

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The Kα X- rays arising from a cobalt (Z=27) target have a wavelength of 179 pm. The Kα X rays arising from a nickle target (Z=28) is

The correct option is B < 179 pm λ∝1(Z−1)2 λNiλCo=(ZCo−1ZNi−1)2=(27−128−1)2 λNi=(26)2(27)2λCo (λNi<λCo)

The $K_{α}$ X- rays arising from a cobalt ( $Z = 27$ ) target have a wavelength of 179 $p m$ . The $K_{α}$ X rays arising from a nickle target (Z=28) is

The correct option is B
< 179 $p m$

$λ \propto \frac{1}{(Z - 1)^{2}}$

$\frac{λ_{N i}}{λ_{C o}} = (\frac{Z_{C o} - 1}{Z_{N i} - 1})^{2} = (\frac{27 - 1}{28 - 1})^{2}$
$λ_{N i} = \frac{(26)^{2}}{(27)^{2}} λ_{C o}$ $(λ_{N i} < λ_{C o})$