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Two hydrogen atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is:
 

Option: 1

10.2 eV


Option: 2

20.4 eV


Option: 3

13.6 eV


Option: 4

27.2 eV


Answers (1)

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Initial kinetic energy of each of two hydrogen atoms in ground state =13.6\; \mathrm{eV}

\therefore \quad   Kinetic energy of both \mathrm{H} atoms before collision =2 \times 13.6 \mathrm{eV}=27.2 \mathrm{eV}

As the collision is inelastic, linear momentum is conserved but some kinetic energy is lost.

If one H atom goes to first excited state and other remains in ground state, then their combined kinetic energy after collision

=\frac{13.6}{2^2} \mathrm{eV}+13.6 \mathrm{eV}=17\; \mathrm{eV}

\therefore \quad   Reduction in their combined kinetic energy =27.2 \mathrm{eV}-17 \mathrm{eV}=10.2 \; \mathrm{eV}

Posted by

Ritika Jonwal

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