Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Help me please, An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wav

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The K.E. of colliding electron will be

 

  • Option 1)

    10.2 eV

     

  • Option 2)

     1.9 eV

     

  • Option 3)

     12.1 eV

  • Option 4)

    13.6 eV

     

 

Answers (1)

best_answer

 

Energy emitted due to transition of electron -

\Delta E= Rhcz^{2}\left ( \frac{1}{n_{f}\, ^{2}}-\frac{1}{n_{i}\, ^{2}} \right )

\frac{1}{\lambda }= Rz^{2}\left ( \frac{-1}{n_{i}\, ^{2}}+\frac{1}{n_{f}\, ^{2}} \right )

- wherein

R= R hydberg\: constant

n_{i}= initial state \\n_{f}= final \: state

 

 Due to collision e-  hydrogen atom gets excited to n=3. hence energy of colliding electron is difference between n=1 and n=3

\therefore E=13.6(1-\frac{1}{9})=12.1ev


Option 1)

10.2 eV

 

Option is incorrect

Option 2)

 1.9 eV

 

Option is incorrect

Option 3)

 12.1 eV

Option is correct

Option 4)

13.6 eV

 

Option is incorrect

Posted by

prateek

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2024 registration window closes tomorrow; exam from May 20
anam-khan

BITSAT 2024 registration deadline extended till April 16; session 1 exam from May 19 to 24

Latest Question