Module 5 : MODERN PHYSICS
Lecture 26 : Wave Nature of Particle - the de Broglie Hypothesis
  Example 16
If an electron makes a transition from $ n=4$ to $ n=2$, determine (i) the wavelength of emitted radiation and (ii) the recoil speed of the electron.
  Solution :
The wavelength of emitted radiation is
 
$\displaystyle \frac{1}{\lambda} = 1.097\times 10^7\left(\frac{1}{4}-\frac{1}{16}\right) = 0.2055\times 10^7\ \ {\rm nm}^{-1}$
  The wavelength is $ \lambda = 486 $ nm. The momentum associated with this radiation is
 
$\displaystyle p = \frac{h}{\lambda} = \frac{6.63\times 10^{-34}}{4.86\times 10^7}= 1.36\times 10^{-27}\ \ {\rm kg- m/s}$
  By conservation of momentum, this is also the magnitude of the momentum imparted to the atom as a whole. The recoil speed of the electron is
 
$\displaystyle v = \frac{p}{m_{atom}}= \frac{1.36\times 10^{-27}}{1.67\times 10^{-27}} = 0.81\ \ {\rm m/s}$

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