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Q.

For  He+ , a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm.
The wavelength (in  A0) of the emitted photon during the transition is  x×102  then  x  
[Use : Bohr radius, a = 52.9 pm; Rydberg constant ,  RH=2.2×1018J  ; Planck’s constant , h=6.6×1034Js  ; Speed of light , c=3×108ms1 ]

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answer is 3.

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Detailed Solution

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For single electron system, rn=52.9×n2Zpm  
105.8=52.9×n122n12=4n1=2
26.45=52.9×n222n2=1
So, transition is from 2 to 1.
Nowhcλ=RHZ2(1n121n22)λ=30×109m=30nm 

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