What is the de Broglie wavelength associated with an electron accelerated through a potential difference of?

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The following graph shown the variation of stopping potential V0 with the frequency v of the incident radiation for two photosensitive metals P and Q:

(i) Explain which metal has smaller threshold wavelengths.(ii) Explain, giving reason, which metal emits photoelectrons having smaller kinetic energy.

(iii) If the distance between the light source and metal P is doubled, how will the stopping potential change? 

(i) Suppose the frequency of incident radiations of metal Q and P be v0 and v0

’ respectively.
                

E = hv0

(iii) Stopping potential remains unaffected because the value of stopping potential for a given metal surface does not depend on the intensity of the incident radiation. It depends on the frequency of incident radiation.

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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.

Let's discuss the concepts related to Atoms and De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation. Explore more from Physics here. Learn now!

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