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Question: The de Broglie Wavelength is given by :...

The de Broglie Wavelength is given by :

A

p=2πhλp = \frac{2\pi h}{\lambda}

B

p=h2λp = \frac{h}{2\lambda}

C

p=2πhλp = \frac{2\pi}{h\lambda}

D

p=2πλp = \frac{2\pi}{\lambda}

Answer

p=2πhλp = \frac{2\pi h}{\lambda}

Explanation

Solution

: de Broglie wavelength is given by λ=hp,\lambda = \frac{h}{p},

Where p is the momentum Or p=hλp = \frac{h}{\lambda}

Since ℏ=h2π∴p=2πhλ\hslash = \frac{h}{2\pi}\therefore p = \frac{2\pi h}{\lambda}