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A hydrogen atom, at rest, emits the β –line of its Balmer series. The recoil energy of the hydrogen atom, in terms of R (= Rydbreg’s constant for hydrogen), h (= Planck’s constant) and M (= Mass of hydrogen atom), would be

(1) \(\frac{9h^2R^2}{256M}\)

(2) \(\frac{9h^2R^2}{512M}\)

(3) \(\frac{25h^ 2R^2}{1296M}\)

(4) \(\frac{25h^2R^2}{512M}\)

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 (2) \(\frac{9h^2R^2}{512M}\)

The β–line, in the Balmer series, corresponds to a transition from n2 = 4 to n1 = 2. The wavelength, λ, of this β–line, is given by

The linear momentum of the emitted photon = p = h/λ or p = 3hR/16

The hydrogen atom is initially at rest. To conserve linear momentum, the recoil momentum of the hydrogen atom, is also p, directed in a direction opposite to the direction of motion of the photon. Hence the recoil energy of the hydrogen atom 

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