Atoms & Nuclei

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Question 27 of 239

The ratio of momentum of the photons of the 1st1^{st} and 2nd2^{nd} line of Balmer series of Hydrogen atoms is α/β\alpha/\beta. The possible values of α\alpha and β\beta are:-

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Answer: (D) 2020 and 2727

The momentum of a photon is given by p=hλp = \dfrac{h}{\lambda}.

From Rydberg's formula, the wave number is 1λ=R(1n12−1n22)\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right).

For the first line of the Balmer series, the transition is from n2=3n_2 = 3 to n1=2n_1 = 2.

p1=hλ1=hR(122−132)=hR(14−19)=5hR36p_1 = \dfrac{h}{\lambda_1} = hR \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) = hR \left( \dfrac{1}{4} - \dfrac{1}{9} \right) = \dfrac{5hR}{36}

For the second line of the Balmer series, the transition is from n2=4n_2 = 4 to n1=2n_1 = 2.

p2=hλ2=hR(122−142)=hR(14−116)=3hR16p_2 = \dfrac{h}{\lambda_2} = hR \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right) = hR \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = \dfrac{3hR}{16}

The ratio of the momenta is:

p1p2=5hR363hR16=536×163=2027\dfrac{p_1}{p_2} = \dfrac{\dfrac{5hR}{36}}{\dfrac{3hR}{16}} = \dfrac{5}{36} \times \dfrac{16}{3} = \dfrac{20}{27}

Comparing with αβ\dfrac{\alpha}{\beta}, the possible values are α=20\alpha = 20 and β=27\beta = 27.

Answer: 2020 and 2727

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