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प्रश्न
A positive ion having just one electron ejects it if a photon of wavelength 228 Å or less is absorbed by it. Identify the ion.
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उत्तर
Given:
228 Å
Energy (E) is given by
`E = (hc)/lamda`
Here, c = Speed of light
h = Planck's constant
`E =((6.63 xx 10^-34) xx (3xx10^8))/(228xx10^-10)`
= 0.0872 × 10-16 J
As the transition takes place from n = 1 to n = 2, the excitation energy (E1) will be
`E_1 = RhcZ^2 (1/(n_1^2 )-1/n_2^2)`
`E_1 = (13.6 eV ) xx Z^2 xx (1/1^2 - 1/2^3)`
`rArr E_1 = (13.6 eV)xxZ^2xx3/4`
This excitation energy should be equal to the energy of the photon.
`therefore 13.6 xx 3/4 xx Z^2 = 0.0872xx10^-16`
`Z^2 = (0.0872xx10^-16xx4)/(13.6xx3xx1.6xx10^-19 )= 5.34`
`Z = sqrt(5.34 = 2.3`
The ion may be helium.
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