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Q.

The wavelength of radiation emitted is λ0 when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the  wavelength of radiation emitted will be

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a

1625 λ0

b

2027 λ0

c

2720 λ0

d

2516 λ0

answer is B.

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Detailed Solution

For transition n = 3 to n = 2,

1λ1=R122-132=5R36               …(i)

For transition n = 4 to n = 2,

1λ2=R122-142=3R16              …(ii)

Now, dividing Eq. (i) by Eq. (ii), we get

λ2λ1=2027 or λ2=2027λ1=2027λ0  λ1=λ0

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