First slide

Bohr's atom model

Question

The minimum kinetic energy required for ionization of a hydrogen atom is E₁ in case electron is collided with hydrogen atom. It is E₂ if the hydrogen ion is collided and E₃ when helium ion is collided. Then,

Moderate

A

$E_{1} = E_{2} = E_{3}$
B

$E_{1} > E_{2} > E_{3}$
C

$E_{1} < E_{2} < E_{3}$
D

$E_{1} > E_{3} > E_{2}$

Solution

Assuming that ionization occurs as a result of a completely inelastic collision, we can write
$m v - 0 = (m + m_{H}) u$
where m is the mass of incident particle, m_H the mass of hydrogen atom, v₀ the initial velocity of incident particle, and u the final common velocity of the particle after collision. Prior to collision, the KE of the incident particle was
$E_{0} = \frac{m v_{0}^{2}}{2}$
The total kinetic energy after collision
$E = \frac{(m + m_{H}) u^{2}}{2} = \frac{m^{2} v_{0}^{2}}{2 (m + m_{H})}$
The decrease in kinetic energy must be equal to ion ionization energy. Therefore,
$E_{1} = E_{0} - E = (\frac{m_{H}}{m + m_{H}}) E_{0}$
$i.e., \frac{E_{1}}{E_{0}} = \frac{1}{1 + \frac{m}{m_{H}}}$
i.e., the greater the mass m, the smaller the fraction of initial kinetic energy that be used for ionization.

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