Question

the de-Broglie wavelength for a proton with a velocity 15% of the speed of light is:​

Answer 1

Answer:

The de-Broglie wavelength for a proton will $8.803\times 10^{-15} m$.

Explanation:

De-Broglie wavelength is calculated by using the formula:

$\lambda=\frac{h}{mv}$      .....(1)

where,

$\lambda$ = wavelength of proton

h = Planck's constant = $6.626\times 10^{-34}Js$

m = mass of proton= $1.6726\times 10^{-27}kg$

v = velocity of proton = 15% of the speed of light :

$v=\frac{15}{100}\times 3\times 10^8=45\times 10^6 m/s$

On substituting the values:

$\lambda =\frac{6.626\times 10^{-34}Js}{1.6726\times 10^{-27}kg\times 45\times 10^6 m/s}=8.803\times 10^{-15} m$

The de-Broglie wavelength for a proton will $8.803\times 10^{-15} m$.