Given X-ray spectrum is for a Coolidge tube having accelerating potential V- If accelerating potential is decreased to V - 4- then - C becomes four times with changes in anode element atomic no- from neglect screening effect-A- Z - 2B- Z-3C- 2 ZD- Z

The correct option is A Z/2 Δλ_1=(λ_1 λ_c^1)=hc/Z^1^2(E_k E_L) hc/eV^1=4(hc/Z^2(E_k E_L) hc/eV )⇒ Z^1=Z/2 ...

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Standard XII

Physics

Hydrogen Spectra

Given X-ray s...

Question

Given X-ray spectrum is for a Coolidge tube having accelerating potential V. If accelerating potential is decreased to V/4, then $Δ λ = (λ - λ_{c})$ becomes four times with changes in anode element atomic no. from (neglect screening effect).

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Z/2

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Z/3

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Solution

The correct option is B
Z/2

$Δ λ_{1} = (λ_{1} - λ_{c}^{1}) = \frac{h c}{Z^{1^{2}} (E_{k} - E_{L})} - \frac{h c}{e V^{1}} = 4 (\frac{h c}{Z^{2} (E_{k} - E_{L})} - \frac{h c}{e V}) \Rightarrow Z^{1} = Z / 2$

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Given X-ray spectrum is for a Coolidge tube having accelerating potential V. If accelerating potential is decreased to V/4, then Δλ=(λ−λc) becomes four times with changes in anode element atomic no. from (neglect screening effect).

The correct option is B Z/2 Δλ1=(λ1−λ1c)=hcZ12(Ek−EL)−hceV1=4(hcZ2(Ek−EL)−hceV)⇒Z1=Z/2

Given X-ray spectrum is for a Coolidge tube having accelerating potential V. If accelerating potential is decreased to V/4, then $Δ λ = (λ - λ_{c})$ becomes four times with changes in anode element atomic no. from (neglect screening effect).

The correct option is B
Z/2

$Δ λ_{1} = (λ_{1} - λ_{c}^{1}) = \frac{h c}{Z^{1^{2}} (E_{k} - E_{L})} - \frac{h c}{e V^{1}} = 4 (\frac{h c}{Z^{2} (E_{k} - E_{L})} - \frac{h c}{e V}) \Rightarrow Z^{1} = Z / 2$