What is the wavelength of the light produced when the electron transition occurs from n 3 to n 2?

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Text Solution

Answer : A::B::C::D

Solution : (a) Reduced mass of positronium and electron is `(m)/(2),` <br> where, m = mass of electron <br> `E prop m :. lambda prop (1)/(m)` <br> m has become half, so `lambda` will become two <br>times or 1312 nm or `1.31 mum.` <br>. For singly ionixed helium atom Z =2 <br> `:. lambda is (1)/(4) th or 164 nm`

Text Solution

Solution : The wavelength of the emitted photon is given by <br> `(1)/(lambda)=R((1)/n_(1)^(2)-(1)/(3^(2)))` <br> When the transition takes place n =3 to n=2, then <br> `(1)/(lambda)=(1.097xx10^(7))((1)/(2^(2))-(1)/(3^(3)))` <br> `=1.097xx10^(7)xx(4)/(36)` <br> `:." " lambda=(36)/(1.097xx10^(7)xx5)=6.563xx10^(-7) m` <br> `=6563Å` <br> `lambda` falsl in the visible (red) part of the specturm, hence the photon will be visible. This photon is the first member of the Balmar series.