The ratio of momentum of the photons of the 1st and 2nd line of Balmer series of Hydrogen atoms is

Question

The ratio of momentum of the photons of the 1st and 2nd line of Balmer series of Hydrogen atoms is $\frac{α}{β}$ . The possible values of $α$ and $β$ are [2026]

Smart Achievers · Accepted Answer

(4)

$The momentum of a photon p = \frac{h}{λ}$

$The Rydberg formula for hydrogen (Z = 1)$

$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

$For the 1^{s t} line (3 \to 2)$

$\frac{1}{λ_{1}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{1}{4} - \frac{1}{9}) = R (\frac{5}{36})$

$For the 2^{n d} line (4 \to 2)$

$\frac{1}{λ_{2}} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = R (\frac{1}{4} - \frac{1}{16}) = R (\frac{3}{16})$

$∴$ $Ratio \frac{P_{1}}{P_{2}} = \frac{1 / λ_{1}}{1 / λ_{2}} = \frac{R (5 / 36)}{R (3 / 16)} = \frac{20}{27}$

$α = 20 and β = 27$

Solution

Q.
The ratio of momentum of the photons of the 1st and 2nd line of Balmer series of Hydrogen atoms is $\frac{α}{β}$ . The possible values of $α$ and $β$ are [2026]

1 27 and 20

2 3 and 16

3 5 and 36

4 20 and 27

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