What is the de Broglie wavelength associated with an electron accelerated through a potential difference of 150 Volt is?

Q: Electrons are accelerated through a potential difference of 150 V. Calculate the de-Broglie wavelength.

Sol: V = 150 V ; h = 6.62 × 10-34 Js, m = 9.1 x 10-31 kg,

e = 1.6 x 10-19 C

$\large \lambda = \frac{h}{\sqrt{2 m e V}} $

$\large \lambda = \frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}\times 150}}$

= 1 A°

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.
                

                 
                 
              Therefore,  metal 'Q' has smaller wavelength.(ii) As we know,

E = hv0

Hence, metal 'P' has smaller kinetc energy.

(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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