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(b) (i) Derive an expression for de Broglie wavelength of an electron.(ii) Find the value of momentum and de Broglie wavelength of particle of mass \( 50 g \) and moving with a speed of \( 200 m / s \).a) Derive the mirror equation. OR |
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Answer» P = mv = 0.050×200 = 10 kgm/s Given m = 50g m = 0.050 kg v = 200m/s (i) Momentum P = mv P = 0.050 x 200 = 50/1000 x 200 P= 10 kg m/s De brogli wavelength \(\lambda = \frac{h}{P}\) \(\lambda = \frac{6.626\times10^{-34}}{1}\) \(\lambda\) = 6.626 x 10-34 m |
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