Question

If an electronic is subjected to a potential of 54 volts than its wavelength is?

Answer 1

Answer:

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Answer 2

Answer:

Using energy conservation, KE=PE or  

2m

p  

2

 

​  

=eV or p=  

2meV

​  

 where p= momentum of electron, m= mass of electron, e= charge of electron and V= potential difference.

Now, de-Broglie wavelength λ=  

p

h

​  

=  

2meV

​  

 

h

​  

=  

2(9.1×10  

−31

)(1.6×10  

−19

)(100)

​  

 

6.6×10  

−34

 

​  

=1.22×10  

−10

m

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Explanation: