Question 133: Dual Nature of Matter and Radiation

Question

The equation $$\lambda=\frac{1.227}{x} \mathrm{~nm}$$ can be used to find the de-Brogli wavelength of an electron. In this equation $$x$$ stands for :
Where
$$\mathrm{m}=$$ mass of electron
$$\mathrm{P}=$$ momentum of electron
$$\mathrm{K}=$$ Kinetic energy of electron
$$\mathrm{V}=$$ Accelerating potential in volts for electron

Accepted Answer

The de Broglie wavelength of a particle can be expressed as:
$$ \lambda = \frac{h}{p} $$
where:
$$ h $$ is Planck's constant, and
$$ p $$ is the momentum of the particle.
For an electron accelerated through a potential difference of $$ V $$ volts, its kinetic energy $$ K $$ is given by:
$$ K = eV $$
where $$ e $$ is the charge of an electron.
The electron's momentum can be expressed in terms of its kinetic energy as:
$$ p = \sqrt{2mK} $$
where $$ m $$ is the mass of the electron.
Substituting this into the de Broglie wavelength equation, we get:
$$ \lambda = \frac{h}{\sqrt{2mK}} $$
Comparing this with the given equation $$ \lambda = \frac{1.227}{x} \, nm $$, we see that $$ x $$ must correspond to:
$$ x = \sqrt{2mK} $$
Since $$ K = eV $$, we can substitute this into our expression for $$ x $$ to get:
$$ x = \sqrt{2meV} = \sqrt{V} $$
where we've used the fact that the mass and charge of an electron are constants.

Answer

$$\sqrt{\mathrm{mK}}$$

Answer

$$\sqrt{\mathrm{P}}$$

Answer

$$\sqrt{\mathrm{K}}$$

Answer

$$\sqrt{\mathrm{V}}$$

Dual Nature of Matter and Radiation

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