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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 932RH
Explanation:
Given:
The wavelength of the emitted photon when an electron from n2 to n1=?

(n2+n1)=4(1)

(n22n21)=8(2)
From equation (1)and (2):

(a2b2)=(a+b)(ab)

(n22n21)=(n2+n1)(n2n1)

(n2+n1)(n2n1)=8

4(n2n1)=8

(n2n1)=2(3)

Now add equation (2) and (3) we get

n2+n1+n2n1=2+4

2n2=6n2=3

n1=4n2

n1=43=1

n1=1,n2=3

From Rydbergs equation we have the formula for wavelength as

1λ=RHZ2[1n211n22]

where,

λ=wavelength=?

Z=atomic number =2

1λ=RH×4[1119]=R×4[919]=RH×4[89]=32RH9

1λ=32RH9

λ=932RH

Hence the correct answer is option (B).



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