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An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V the Value of n is. 

  • Option 1)

    3

  • Option 2)

    4

  • Option 3)

    5

  • Option 4)

    2

 

Answers (1)

best_answer

As we learnt in 

Lymen series -

frac{1}{lambda }= Rleft ( frac{1}{1^{2}}- frac{1}{n^{2}} 
ight )

left ( uv : region 
ight )

- wherein

n=2,3,4,-----

When electron jump from higher orbital to 1st energy level

i.e. ground state

    

 

 Energy of photon emitted from the hydrogen atom is

\\ E=\frac {1}{2}m\upsilon ^{2}+\phi \\ \\ E=eV_{s}+\phi \\ \\ V_{s}=stopping \ potential \\ \\ \therefore \: E=10+2.75=12.75 eV

This is equal to energy gap between n=1 and n=4 in hydrogen atom.

 


Option 1)

3

Incorrect

Option 2)

4

Correct

Option 3)

5

Incorrect

Option 4)

2

Incorrect

Posted by

Plabita

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