An electron collides with a Hydrogen atom in its ground state and excites it to $n = 3$ state. The energy given to the Hydrogen atom in his inelastic collision (neglecting the recoil of Hydrogen atom) is

10.2 e V

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12.1 e V

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12.5 e V

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13.6 e V

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Solution

The correct option is A $12.1 e V$
Energy given to hydrogen atom $=$ difference in energy levels from $n_{1} = 1$ to $n_{2} = 3$ states
Energy difference $Δ E$ $= 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$ eV
$Δ E = 13.6 (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = 12.1 e V$

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An electron collides with a Hydrogen atom in its ground state and excites it to n=3 state. The energy given to the Hydrogen atom in his inelastic collision (neglecting the recoil of Hydrogen atom) is

The correct option is A 12.1eVEnergy given to hydrogen atom = difference in energy levels from n1=1 to n2=3 statesEnergy difference ΔE =13.6(1n21−1n22) eVΔE=13.6(112−132)=12.1eV

An electron collides with a Hydrogen atom in its ground state and excites it to $n = 3$ state. The energy given to the Hydrogen atom in his inelastic collision (neglecting the recoil of Hydrogen atom) is

The correct option is A $12.1 e V$
Energy given to hydrogen atom $=$ difference in energy levels from $n_{1} = 1$ to $n_{2} = 3$ states
Energy difference $Δ E$ $= 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$ eV
$Δ E = 13.6 (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = 12.1 e V$