The recoil speed of a free H-atom at rest in fourth excited state- when it undergoes a transition to ground state is mass of the H atom -1-67 - 10-27 kg and assume exciting energy is taken by the atomwrite upto two decimal places in m - s is -Take h -6-68 - 10-34 Js and -R H -1 - 107 m -1-

As given: nbsp; mass of the H atom = 1.67 × 10^ 27 kg c=3 × 10^8 m/s Energy released during transition: E= hcR_H(1/n_1^2 1/n_2^2) Momentum of the photon = E/c ...

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Byju's Answer

Standard XII

Physics

Oblique Collision

The recoil sp...

Question

The recoil speed of a free H-atom at rest in fourth excited state, when it undergoes a transition to ground state is (mass of the H atom = $1.67 \times 10^{- 27}$ kg and assume exciting energy is taken by the atom)(write upto two decimal places) in $m / s$ is
[Take $h = 6.68 \times 10^{- 34} J s$ and $R_{H} = 1 \times 10^{7} m^{- 1}$ ]

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Solution

As given:
mass of the H-atom = $1.67 \times 10^{- 27}$ kg
$c = 3 \times 10^{8}$ m/s
Energy released during transition: $E = h c R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Momentum of the photon = $\frac{E}{c} = h R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

From principle of linear momentum $m v = h R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$\Rightarrow v = 3.84 m / s$

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The recoil speed of a free H-atom at rest in fourth excited state, when it undergoes a transition to ground state is (mass of the H atom = 1.67×10−27 kg and assume exciting energy is taken by the atom)(write upto two decimal places) in m/s is [Take h=6.68×10−34 Js and RH=1×107 m−1]

As given: mass of the H-atom = 1.67×10−27 kg c=3×108 m/s Energy released during transition: E=hcRH(1n21−1n22) Momentum of the photon = Ec=hRH(1n21−1n22) From principle of linear momentum mv=hRH(1n21−1n22) ⇒v=3.84 m/s