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

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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 $


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