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Question

The value of (n1+n2) and (n22−n21) for He+ ion in atomic spectrum are 4 and 8 respectively. The wavelength of emitted radiation when electrons jump from n2 to n1 is:

A
329RH
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B
932RH
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C
329RH
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D
932RH
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Solution

The correct option is B 932RHExplanation:Given:The wavelength of the emitted photon when an electron from n2 to n1=?(n2+n1)=4→(1)(n22−n21)=8→(2)From equation (1)and (2):∵(a2−b2)=(a+b)(a−b)(n22−n21)=(n2+n1)(n2−n1)(n2+n1)(n2−n1)=84(n2−n1)=8(n2−n1)=2→(3)Now add equation (2) and (3) we getn2+n1+n2−n1=2+42n2=6⇒n2=3n1=4−n2n1=4−3=1∴n1=1,n2=3From Rydberg′s equation we have the formula for wavelength as1λ=RHZ2[1n211n22]where,λ=wavelength=?Z=atomic number =2 ⇒1λ=RH×4[11−19]=R×4[9−19]=RH×4[89]=32RH9⇒1λ=32RH9∴λ=932RHHence the correct answer is option (B).

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