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Questions  >>  JEEMAIN and NEET  >>  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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