Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The binding energy of an electron in the ground state of He is equal to . The energy required to remove both the electrons

The binding energy of an electron in the ground state of He is equal to 24.6 \mathrm{eV}. The energy required to remove both the electrons is:
 

Option: 1

46.2 \mathrm{eV}


Option: 2

21.6 \mathrm{eV}


Option: 3

38.5 \mathrm{eV}


Option: 4

79.0 \mathrm{eV}


Answers (1)

best_answer

Helium atom has 2 electrons. When one electron is removed, the remaining atom is hydrogen like atom, whose energy in first orbit is

\mathrm{E}_1=-(2)^2(13.6 \mathrm{eV})=-54.4 \mathrm{eV}

Therefore, to remove the second electron from the atom an additional energy of 54.4 \mathrm{eV} is required.

Hence, total energy required to remove both the electrons =24.6+54.4=79.0 \mathrm{eV}.

Posted by

Rishi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question