A moving Hydrogen atom makes a head-on collis | vedclass

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Question

(C) Let the mass of each Hydrogen atom be $m$. Let the initial velocity of the moving atom be $v$ and the stationary atom be $0$. After the collision,

Vedclass · Accepted Answer

$20.4$

Vedclass · Answer

(C) Let the mass of each Hydrogen atom be $m$. Let the initial velocity of the moving atom be $v$ and the stationary atom be $0$. After the collision,they move together with velocity $V$. By conservation of momentum: $mv = (m+m)V \implies V = v/2$. The kinetic energy lost in the inelastic collision is $\Delta K = K_i - K_f = \frac{1}{2}mv^2 - \frac{1}{2}(2m)V^2 = \frac{1}{2}mv^2 - \frac{1}{2}(2m)(v/2)^2 = \frac{1}{2}mv^2 - \frac{1}{4}mv^2 = \frac{1}{4}mv^2$. This lost energy is used to excite one of the atoms to the first excited state $(n=2)$. The energy required for the first excitation is $\Delta E = E_2 - E_1 = -3.4 \ eV - (-13.6 \ eV) = 10.2 \ eV$. Setting the lost energy equal to the excitation energy: $\frac{1}{4}mv^2 = 10.2 \ eV$. Therefore,the initial kinetic energy $K_i = \frac{1}{2}mv^2 = 2 	imes 10.2 \ eV = 20.4 \ eV$.

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