_{0}c^{2 }& m_{p} = 1840m_{e} )

given :

kinetic energy = rest mass energy

or

K = m_{0}c^{2}

so, deBroglie wavelength will be

λ = h / (2m_{0}K)^{1/2}

or

= h / (2m_{0}.m_{0}c^{2})^{1/2}

so,

λ = h / (2m_{0}^{2}c^{2})^{1/2} = h / (m_{0}c√2)

here

h = 6.63 x 10^{-34} J.s

m_{0} = 1840m_{e} = 1840 x 9.1 x 10^{-31} kg = 1.67 x 10^{-27} kg

c = 3.8 x 10^{8} m/s

thus,

λ = 6.63 x 10^{-34} / [ 1.67 x 10^{-27} x 3.8 x 10^{8} x √2]

so, the wavelength of the proton will be

**λ = 0.738 x 10**^{-15}** m**

**
**