Question

The equation $\lambda=\frac{1227}{ x } nm$ can be used to find the de-Brogli wavelength of an electron In this equation $x$ stands for : Where, $m =$ mass of electron $P =$ momentum of electron $K =$ Kinetic energy of electron $V=$ Accelerating potential in volts for electron

• A. $\sqrt{m K}$
• B. $\sqrt{ P }$
• C. $\sqrt{ K }$
• D. $\sqrt{ V }$

Answer 1

The Correct Option is D

The equation given is \lambda=\frac{1227}{x} \text{ nm}, which is used to determine the de Broglie wavelength of an electron. Let us decipher what each component represents and find out what x stands for.

The de Broglie wavelength \lambda of a particle such as an electron can be expressed using its momentum or kinetic energy. The fundamental de Broglie relation is:

\lambda = \frac{h}{p}

Where:

• \lambda is the wavelength,
• h is Planck's constant, and
• p is the momentum of the particle.

However, when an electron is accelerated by an electric potential V, its kinetic energy K is given by:

K = eV

This kinetic energy can be related to the momentum p by:

p = \sqrt{2mK}

Substituting K = eV into the momentum equation gives:

p = \sqrt{2meV}

Replacing p in the de Broglie equation gives us:

\lambda = \frac{h}{\sqrt{2meV}}

Now, simplifying for the factor that appears in the denominator:

If h/\sqrt{2m \text{ constant } V} = \frac{1227}{\sqrt{V}} \text{ nm}, it is clear that x is connected to the potential V as the square root.

Thus, in the given equation, x = \sqrt{V}.

Therefore, the correct answer is \sqrt{V}, which corresponds to the accelerating potential in volts for the electron.