Saved Bookmarks
| 1. |
A hydrogen like atom with atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition ot quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is – 13. 6 eV.A. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - B Let ground state energy (in eV) be `E_(1)` Then from the given condition `E_(2n)-E_(1)=204eV` or `(E_(1))/(4n^(2))-E_(1)=204eV` `rArr E_(1)((1)/(4n^(2))-1)=204eV`...(i) and `E_(2n) -E_(n) = 40.8 eV` `rArr (E_(1))/(4n^(2))-(E_(1))/(n^(2))=E_(1)(-(3)/(4n^(2)))=40.8eV`....(ii) From equation (i) and (ii), `(1-(1)/(4n^(2)))/((3)/(4n^(2)))=5 rArr n=2` . |
|