The emission series of hydrogen atom is given by 1 =R ( 1 n₁² - 1 n₂² ) where R is the Rydberg constant. For a transition from n ₂ to n ₁ , the relative change / in the emission wavelength if hydrogen is replaced by deuterium (assume that the mass of proton and neutron are the same and approximately 2000 times larger than that of electrons) is

Step-by-step solution

R = m _ e e ⁴ 8 ₀² ch ³ where m _ e = mass of electron When we consider mass of nucleus also then we replace m with reduced mass = m_ e m m_ e +m where m = mass of nucleus In case of hydrogen atom ₁= ( m _ e ) (2000 ~m _ e ) m _ e +(2000) m _ e = 2000 2001 ~m _ e In case of deuterium ₂= ( m _ e ) (4000 ~m _ e ) m _ e +(4000) m _ e = 4000 4001 ~m _ e Hence, aligned ₂ ₁ &= R ₂ R ₁ = ₂ ₁ &= ( 4000 ~m _ e 4001 ) (2001) (2000 ~m _ e ) &= 4002 4001 ₂ ^ - ₁ ₁ =& 1 4001 ₁ 100 &= 1 4001 100 &=0.025 % aligned