Atoms And Nuclei Question 328
Question: If radiation corresponding to first line of “Balmer series” of $ H{{e}^{+}} $ ion knocked out electron from 1st excited state of H atom, the kinetic energy of ejected electron from H atom would be (eV) - [Given $ E_{n}=-\frac{Z^{2}}{n^{2}}( 13.6eV ) $ ]
Options:
A) 4.155 eV
B) 8.310 eV
C) 2.515 eV
D) 5.550 eV
Show Answer
Answer:
Correct Answer: A
Solution:
- Energy of photon corresponding to first line of Balmer series $ =( 13.6 ){{( 2 )}^{2}}[ \frac{1}{4}-\frac{1}{9} ] $ Energy need to eject electron from n=2 level in H atom $ =( 13.6 )( \frac{1}{4} ) $ So, required kinetic energy $ =( 13.6 )[ ( \frac{1}{4}-\frac{1}{9} )-( \frac{1}{4} ) ]eV $ $ =13.6\times ( \frac{11}{36} )=4.155eV $
BETA
