Binding energy of electron MCQs with Answers

Binding energy of electron MCQs

According to Bohr’s postulates when an electron revolves around the nucleus, the centripetal force required for the circular motion of electron is provided by electrostatic force of attraction between the negatively charged electron and the positively charged nucleus. So that, the energy of electron is inversely proportional to square of principal quantum number. The formula for the energy by which electron in nth orbit is bonded with nucleus is as,

Now to remove electron from the orbit we need to provide the same amount of energy in opposite direction. The minimum energy required to make electron free from the nucleus, is called Binding energy of electron.

This energy is also called ionization energy of hydrogen atom.

Q.1) The minimum energy required by electrons to escape from the orbit is known as……..

a) kinetic energy

b) potential energy

c) gravitational potential energy

d) binding energy

Q.2) The binding energy required for removing the electron from 1st Bohr’s orbit is…..

a) -13.6 eV

b) +13.6 eV

c) -27.2 eV

d) +27.2 eV

Q.3) Binding energy of an electron is…….

a) inversely proportional to n2

b) inversely proportional to ϵ2

c) directly proportional to n2

d) directly proportional to n2

Q.4) The ratio of Binding energy of an electron to 3rd to 1st Bohr’s orbit is….

a) 1.36 × 10-19 J

b) -36 × 10-19 J

c) 1.36 × 10-19 eV

d) -1.36 × 10-19 eV

Q.6) Which among the following is the correct formula for finding binding energy in given orbit?

 

 

 

 

 

Q.7) Difference in the binding energy of electron in 2nd orbit to that of 4th orbit is….

a) 0.55 eV

b) 1.55 eV

c) -0.55 eV

d) -1.55 eV

 

Answers:

Q.1) d) binding energy

Q.2) b) +13.6 eV

Q.3) a) inversely proportional to n2

Q.4) d) 9:1

Q.5) a) 1.36 × 10-19 J

Q.6) a)

Q.7) b) 1.55 eV


Updated: July 5, 2021 — 1:04 pm

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