An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom is excited and colliding electron loses of its kinetic energy. Consequently, the hydrogen atom may emit a photon corresponding to largest wavelength of the Balmer series. Minimum possible kinetic energy of colliding electron will be

10.2 eV

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1.9 eV

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12.1 eV

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1.3 eV

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Solution

The correct option is C 12.1 eV
Largent wavelengths of Balmer series corresponds to n = 3 to n = 2 transition.
Thus, the incoming electron must have an energy
= energy required to reach n = 3 state
$= (\frac{- 13.6}{3^{2}} - (- 13.6))$
=12.09 = 12.1 eV

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The correct option is C 12.1 eVLargent wavelengths of Balmer series corresponds to n = 3 to n = 2 transition. Thus, the incoming electron must have an energy = energy required to reach n = 3 state =(−13.632−(−13.6)) =12.09 = 12.1 eV

The correct option is C 12.1 eV
Largent wavelengths of Balmer series corresponds to n = 3 to n = 2 transition.
Thus, the incoming electron must have an energy
= energy required to reach n = 3 state
$= (\frac{- 13.6}{3^{2}} - (- 13.6))$
=12.09 = 12.1 eV