A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency v 1 and ejects the electron with a kinetic energy of 10 eV . The electron then combines with a positron at rest

Question

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $v_{1}$ and ejects the electron with a kinetic energy of 10 eV . The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $v_{2}$ . The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV . It is given that positron has the same mass as that of electron and the positronium atom can be considered as a Bohr atom, in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process, the difference between the two photon energies (in eV ) is

Question diagram: A hydrogen atom, initially at rest in its ground state, abso

Answer:

11.80

Solution:

$h v_{1} = 13.6 + 10 = 23.6 eV .... (1)$ Energy of positronium in ground state $= - 13.6 \frac{μ}{m} (\frac{z}{n})^{2} eV = - 13.6 \times \frac{1}{2} eV = - 6.8 eV$ So to make positronium 6.8 eV must release \& 5 eV is the KE of COM . So total energy of photon released $(h v_{2})$ will be : $h v_{2} = (10 - 5) + 6.8 = 11.8 eV .... (2)$ $∴$ Difference in energy $= 23.6 - 11.8 = 11.8 eV$

TestHub · Accepted Answer

h v 1 ​ = 13.6 + 10 = 23.6 eV .... ( 1 ) Energy of positronium in ground state ​ = − 13.6 m μ ​ ( n z ​ ) 2 eV = − 13.6 × 2 1 ​ eV = − 6.8 eV ​ So to make positronium 6.8 eV must release \& 5 eV is the KE of COM . So total energy of photon released ( h v 2 ​ ) will be : h v 2 ​ = ( 10 − 5 ) + 6.8 = 11.8 eV .... ( 2 ) ∴ Difference in energy = 23.6 − 11.8 = 11.8 eV

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