Question

Calculate the wavelength of de–Broglie waves associated with a proton of kinetic energy 500eV. (given : mp = $1.67 \times 10^{-27}$kg, h = $6.626 \times 10^{-34}$Js and 1eV = $1.6 \times 10^{-19}$J)

Answer 1

To determine:  The de-Broglie wavelength of a proton possessing kinetic energy of 500 eV

Given Data: Mass of the proton $= 1.67 \times 10^{-27} kg$

                  $1\quad eV\quad =\quad 1.6\quad \times \quad 10^{ -19 }$

Formulas to be used:

De-Broglie's wavelength $=\quad \lambda \quad =\quad \frac { h }{ p }$

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

                  $p \rightarrow Momentum$

Kinetic energy of a proton $=\quad E\quad =\quad \frac { 1 }{ 2 } m{ v }^{ 2 }$

Where $m \rightarrow$ mass of the proton

$v \rightarrow$ velocity of the proton

Momentum, $p\quad =\quad mv$

Calculation:

Step 1: Find the momentum using the formula of kinetic energy

                 $p\quad =\quad mv$

                 $E\quad =\quad \frac { 1 }{ 2 } m{ v }^{ 2 }\\V\quad =\sqrt { \frac { 2E }{ m }  }$

So, $p\quad =\quad m\quad \times \sqrt { \frac { 2E }{ m }  } \quad =\quad \sqrt { 2Em }$

Step 2:  Substitute all the necessary values and find the de-Broglie's wavelength

                 $\lambda \quad =\quad \frac { h }{ p }$

                 $=\quad \frac { h }{ \sqrt { 2Em }  }$

                 $=\quad \frac { 6.626\quad \times \quad 10^{ -34 } }{ \sqrt { 2\quad \times \quad 500\quad \times \quad 1.6\quad \times \quad 10^{ -19 }\quad \times \quad 1.67\quad \times \quad 10^{ -27 } }  }$

                 $\lambda \quad =\quad 1.3\quad \times \quad 10^{ -12 }\quad m$

Answer 2

Answer: 1.3×10^-12m

Explanation: First, from formula of kinetic energy(K.E=1/2 m×v^2), we find the momentum(p=mv) of the proton in terms of kinetic energy and mass.

Then by applying the deBroglie's hypothesis(lambda=h/p), we find the wave length of the given proton.

Quite easy right