Question

Consider two moving particles X and Y. the de Brogile wavelength of X is 60 nm. calculation the de brogile wavelength of Y if its momentum is half that of X.​

Answer 1

Answer :

• de Brogile wavelength of Y = 120 nm.

Step-by-step explanation :

Given that,

• de Brogile wavelength of X = 60 nm.
• Momentum of Y is half of that of X.

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Suppose $\rm{{\lambda}_x}$ is the de Brogile wavelength of X and $\rm{P_x}$ is its momentum, then,

$\rm{{\lambda}_x=\dfrac{h}{P_x}=60\:nm\longrightarrow{}(1)}$

$\rm{Now,\:{\lambda}_y=\dfrac{h}{P_y}=\dfrac{h}{P_x/2}\quad\left({\because}\: P_y=\dfrac{P_x}{2}\right)}$

$\implies\rm{{\lambda}_y=\dfrac{2h}{P_x}=2\left(\dfrac{h}{P_x}\right)}$

Using (1), we get,

$\implies\rm{{\lambda}_y=2\times{}60=}\:\bold{120\:nm.}$