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Help me solve this Two particles move at right angle to each other de Broglie wavelengths as and , particle suffers a perfectly inelastic collision. The de Broglie wavelength of final particle is given by

Two particles move at right angle to each other de Broglie wavelengths as and , particle suffers a perfectly inelastic collision. The de Broglie wavelength of final particle is given by

• Option 1)

$\lambda=\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$

• Option 2)

$\lambda=2\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$

• Option 3)

$\lambda=\frac{\lambda _{1}^2\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$

• Option 4)

$\lambda=\frac{\lambda _{1}\lambda _{2}}{2\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$

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option 3

Option 1)

$\lambda=\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$

Option 2)

$\lambda=2\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$

Option 3)

$\lambda=\frac{\lambda _{1}^2\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$

Option 4)

$\lambda=\frac{\lambda _{1}\lambda _{2}}{2\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$

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