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For the hydrogen atom E(n)=-1/(n^(2))R(H...

For the hydrogen atom `E_(n)=-1/(n^(2))R_(H)`
Where `R_(H)=2.178xx10^(-18)J`
Assuming that the electrons in other shells exert no effect, find the minimum no. of Z (atomit no.) at which a transition from the second energy level to the first results in teh emission of an X-ray. Given that the lowest energy rays have `lamda=4xx10^(-8)m`.

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To solve the problem, we will follow these steps: ### Step 1: Calculate the energy difference (ΔE) between the second and first energy levels of the hydrogen atom. The energy levels for a hydrogen atom are given by the formula: \[ E_n = -\frac{R_H}{n^2} \] where \( R_H = 2.178 \times 10^{-18} \, J \). ...
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