What is the de Broglie wavelength of an electron after being accelerated through A potential difference of 54 kv?

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CONCEPT:

• de-Broglie wavelength:  The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. 
• The de-Broglie wavelength of an electron is given by 

λ = h/ √ 2mev

• here. λ is the wavelength of an electron.

h = 6.63 x 10-34

me = mass of an electron

v = potential difference

EXPLANATIONS:

we know 

Plancks constant (h) = 6.63 x 10-34.

Volt = 100 v

Mass of electron (me ) = 9.1 x 10-31..

Putting all these values in  λ = h/√ 2mev

6.63 x 10-34/√ 2 x 9.1 x 10-31  x  100

0.123 nm

Option 1 is correct.

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