Question

A proton with an energy of 2.05Mev has a de -broglie wavelength of

Answer 1

A solution in which the maximum amount of solvent has been dissolved. Any more solute added will sit as crystals on the bottom of the container.

Answer 2

Given:

Energy of the proton $=2.05MeV$

To find:

De-Broglie wavelength of the proton

Solution:

Step 1

In photoelectric effect, the kinetic energy of a particle is numerically equal to its total energy.

                           $K=E$

Kinetic energy of an object is the energy possessed by the object  relative to its velocity.

$KE=\frac{1}{2}mv^{2}$

Multiplying and dividing the equation by mass m, we get

$KE=\frac{1}{2}mv^{2}\frac{m}{m}$

$KE=\frac{1}{2m}m^{2} v^{2}$

We know, momentum is given by $P=mv$

Hence,

$K=\frac{P^{2} }{2m}$       $;$ and

$P=\sqrt{2mK}$

Step 2

We also know,

Einstein's energy mass equivalence says that $E=mc^{2}$

Planck's energy equivalence $E=hf$

Equating both energy relations, we get

$mc^{2} =hf$

$mc^{2} =hc$$/$λ

λ $=\frac{h}{mc}$

λ $=\frac{h}{P}$

This is called as the de-Broglie wavelength.

From $(i)$, we get

λ $=\frac{h}{\sqrt{2mK} }$  

λ$=\frac{h}{\sqrt{2mE} }$

Solution:

We have,

$h=6.6$ × $10^{-34}Js$    $;$  $m=1.6$×$10^{-27}kg$    $;$

$E=2.05MeV$ $=2.05$ × $10^{6}$ × $1.6$ × $10^{-19}J$

Substituting the given values in the equation, we get

λ$=\frac{6.6*10^{-34} }{2(1.6*10^{-27})(2.05*10^{6}*1.6*10^{-19} )}$

λ$=\frac{6.6*10^{-34} }{10.496*10^{-27-19+6} }$

λ $=0.628$ × $10^{40-34}$

λ $=6.28$ × $10^{5}m$

Final answer:

Hence, the de-Broglie wavelength of the proton is 6.28 × 10⁵ m.