A neutral hydrogen molecule has two protons and two electrons. If one of the electrons is removed we get a hydrogen molecular ion $(H_{2}^{+})$ . In the ground state of $H_{2}^{+}$ the two protons are separated by roughly $1.5$ $A^{o}$ and the electrons is roughly $1$ $A^{o}$ from each proton. The potential energy of the system is :

- 38.4 e V

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- 19.2 e V

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- 9.6 e V

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Solution

The correct option is B $- 19.2 e V$
The energy of the system is given by,
$k e^{2} / (1.5 a n g s t r o m) + 2 (k (- e^{2}) / (1 a n g s t r o m))$
$= 9 \times 10^{9} \times 10^{10} \times (2 e^{2} / 3 - 2 e^{2})$
$= 9 \times 10^{19} \times (- 4 / 3) e e V = - 19.2 e V$

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Q. An electron and a proton are separated by a large distance and the electron approaches the proton with a kinetic energy of 4.11 eV. If the electron is captured by the proton to form a hydrogen atom in the ground state. the wavelength of photon given off is

α \times 10^{2} A^{o}

. What is the value of

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The correct option is B −19.2 eVThe energy of the system is given by,ke2/(1.5angstrom)+2(k(−e2)/(1angstrom))=9×109×1010×(2e2/3−2e2)=9×1019×(−4/3)eeV=−19.2eV

The correct option is B $- 19.2 e V$
The energy of the system is given by,
$k e^{2} / (1.5 a n g s t r o m) + 2 (k (- e^{2}) / (1 a n g s t r o m))$
$= 9 \times 10^{9} \times 10^{10} \times (2 e^{2} / 3 - 2 e^{2})$
$= 9 \times 10^{19} \times (- 4 / 3) e e V = - 19.2 e V$