Question

A proton is moving with kinetic energy 5*10^-27j what is the wavelength of the de broglie wave associated with it

Answer 1

Answer:

lambda=1.62×10^-7

Explanation:

mass of proton=m=1.67×10^-27

kinetic energy=1/2mv^2

5×10^-27=(1/2)×1.67×10^-27×v^2

v=2.44

Lambda=h/mv

lambda=6.63×10^-34/(1.67×10^-27×2.44)

lambda=1.62×10^-7

Answer 2

The De- Broglie wavelength is $1.62\times 10^{-7}\ m$

Explanation:

Kinetic energy of the proton is $5\times 10^{-27}\ J$. It is required to find the wavelength of the De Broglie wave associated with it. The kinetic energy of proton is given by :

$E=\dfrac{1}{2}mv^2$

m is the mass of proton

$v=\sqrt{\dfrac{2E}{m}} \\\\v=\sqrt{\dfrac{2\times 5\times 10^{-27}}{1.67\times 10^{-27}}} \\\\v=2.44\ m/s$

The De Broglie wavelength is given by :

$\lambda=\dfrac{h}{mv}\\\\\lambda=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 2.44}\\\\\lambda=1.62\times 10^{-7}\ m$

So, the De- Broglie wavelength is $1.62\times 10^{-7}\ m$.

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De Broglie wavelength

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