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. Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now |