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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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A monochromatic radiation wavelength $\;\lambda_{1}\;$ is incident on a stationary atom as a result of which wavelength of photon after collision becomes $\;\lambda_{2}\;$ and recoiled atom has De- Broglie's wavelength $\;\lambda_{3}\;$.Then

$(a)\;\lambda_{3}=\sqrt{\lambda_{1} \lambda_{2}}\qquad(b)\;\lambda_{1}=\large\frac{\lambda_{2} \lambda_{3}}{\lambda_{2} +\lambda_{3}}\qquad(c)\;\lambda_{1}=\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}\qquad(d)\;\lambda_{3}=\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}$

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1 Answer

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Answer : (b) $\;\lambda_{1}=\large\frac{\lambda_{2} \lambda_{3}}{\lambda_{2} + \lambda _{3} }$
Explanation :
Conservation of momentum yields
$\large\frac{h}{\lambda_{1}}+0=\large\frac{h}{\lambda_{2}}+mV$
$\large\frac{h}{\lambda_{1}}-\large\frac{h}{\lambda_{2}}=mV$
$\large\frac{1}{\lambda_{1}}-\large\frac{1}{\lambda_{2}}=\large\frac{mV}{h}-----(1)$
$\large\frac{h}{mV}=\lambda_{3}$
$\large\frac{1}{\lambda_{1}}-\large\frac{1}{\lambda_{2}}=\large\frac{1}{\lambda_{3}}$
$\large\frac{1}{\lambda_{1}}=\large\frac{1}{\lambda_{2}}+\large\frac{1}{\lambda_{3}}$
$\lambda_{1}=\large\frac{\lambda_{2} \lambda_{3}}{\lambda_{2} + \lambda _{3} }\;.$
answered Feb 25, 2014 by yamini.v
 

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