Question

find momentum of thee electron having de broglie wavelength of 0.5 A

Answer 1

Given:

De- Broglie Wavelength $=0.5$ Angstrom.

To find:

Momentum of the electron.

Solution:

Step 1

According to Planck's equation of energy,

$E=hf$  

Where, $h$ is Planck's constant ( $h=6.6$ × $10^{-34}Js$) and $f$ is frequency.

Also, Einstein's mass energy equivalence states that

$E=mc^{2}$

Where, $c$ is the speed of light in vacuum ( $c=3$ × $10^{8}m/s$ ) and $m$ is mass.

Equating both the relations of energy, we get

$hf=mc^{2}$

We know, $c=f$ × λ, therefore, we get,

$hf=m$ × $c$ × $c$

$hf=mc$ × $f$ × λ

$h=mc$ × λ

λ $=\frac{h}{mc}$

We know, $mc$ is momentum, denoted by $P$.

λ $=\frac{h}{P}$

This is called as the De - Broglie Wavelength.

Step 2

Now,

According to the question, we have

λ $=0.5$ × $10^{-10}m$

Substituting the known values in the equation, we get

$0.5*10^{-10} =\frac{6.6*10^{-34} }{P}$

$P=13.2$ × $10^{-24}kgm/s$

Final answer:

Hence, the momentum will be 13.2 × 10⁻²⁴ kg m/s.