a hydrogen like atom has atomic number Z when it collides with stationary hydrogen atom in ground state it needs a minimum kinetic energy of 51eV to excited hydrogen atom
Question
a hydrogen like atom has atomic number Z when it collides with stationary hydrogen atom in ground state it needs a minimum kinetic energy of 51eV to excited hydrogen atom
Solution
To answer the question, we need to consider the energy required to excite a hydrogen atom when it collides with another hydrogen atom in its ground state.
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The given information states that the hydrogen-like atom has an atomic number Z. This means that it has Z protons in its nucleus, similar to hydrogen.
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When the hydrogen-like atom collides with a stationary hydrogen atom in its ground state, it needs a minimum kinetic energy of 51eV to excite the hydrogen atom.
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The minimum kinetic energy required to excite an atom can be calculated using the formula:
KE = E_excitation - E_ground
where KE is the kinetic energy, E_excitation is the energy of the excited state, and E_ground is the energy of the ground state.
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In this case, the energy of the ground state of hydrogen is known to be -13.6 eV. Therefore, the energy of the excited state can be calculated as:
E_excitation = E_ground + KE
E_excitation = -13.6 eV + 51 eV
E_excitation = 37.4 eV
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Hence, the minimum energy required to excite the hydrogen atom is 37.4 eV.
Please note that the calculation assumes that the hydrogen-like atom and the hydrogen atom are in a one-dimensional collision and that there are no other factors affecting the energy transfer.
Similar Questions
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