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[" For emission line of atomic hydrogen from "n_(i)=8" to "n_(f)=n" ,"],[" the plot of wave number "(bar(v))" against "((1)/(n^(2)))" will be (The "],[" Rydberg constant,"R_(H)" is in wave number uint) "],[" (1) Linear with intercept "-R_(H)],[" (2) Non linear "],[" (3) Linear with slope "R_(H)],[" (4) Linear with slope "-R_(H)]

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For emission line of atomic hydrogen from n_(i)=8 to n_(f)=n, the plot of wave number (barv) against ((1)/(n^(2))) will be (The Rydberg constant, R_(H) is in wave number unit)

For emission line of atomic hydrogen from n_(i)=8 to n_(f)=n, the plot of wave number (barv) against ((1)/(n^(2))) will be (The Rydberg constant, R_(H) is in wave number unit)

For emission line of atomic hydrogen from n_(i)=8 to n_(f)=n, the plot of wave number (barv) against ((1)/(n^(2))) will be (The Rydberg constant, R_(H) is in wave number unit)

For emission line of atomic hydrogen from n_(i)=8 to n_(f)=n, the plot of wave number (barv) against ((1)/(n^(2))) will be (The Rydberg constant, R_(H) is in wave number unit) (1) Linear with slope - RH (2) Linear with intercept-RH (3) Non linear (4) Linear with slope RH

If in Hydrogen atom, an electron jumps from n_(2) = 2 to n_(1) = 1 in Bohr.r orbit, then the value of wave number of the emitted photon will be (R = 109700 cm^(-1))

The wave number of first line in Balmer series of hydrogen spectrum is (Rydberg constant, R_H = 109,678 cm^-1 ) nearly

When a transition of electron in He^(+) takes place from n_(2)" to "n_(1) then wave number in terms of Rydberg constant R will be ("Given "n_(1)+n_(2)=4, n_(2)-n_(1)=2)

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