Question 105: Dual Nature of Matter and Radiation

Question

The de Broglie wavelength of an electron having kinetic energy $\mathrm{E}$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its de-Broglie wavelength will be :

Accepted Answer

$$
\lambda = \frac{h}{\sqrt{2mE}}
$$
where $h$ is Planck's constant, $m$ is the mass of the particle, and $E$ is its kinetic energy.
We are given that the de Broglie wavelength of an electron with kinetic energy $E$ is $\lambda$, and we want to find the de Broglie wavelength of the same electron when its kinetic energy becomes $\frac{E}{4}$.
To do this, we can use the formula for the de Broglie wavelength again, but with the new kinetic energy $\frac{E}{4}$:
$$
\lambda' = \frac{h}{\sqrt{2m\left(\frac{E}{4}\right)}} = \frac{2h}{\sqrt{2mE}} = 2\lambda
$$
where we have used the fact that $\sqrt{\frac{1}{4}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$ to simplify the expression.
Therefore, the de Broglie wavelength of the electron when its kinetic energy becomes $\frac{E}{4}$ is twice its original value, or $\boxed{2\lambda}$.

Answer

$\sqrt{2} \lambda$

Answer

$2 \lambda$

Answer

$\frac{\lambda}{2}$

Answer

$\frac{\lambda}{\sqrt{2}}$

Dual Nature of Matter and Radiation

The de Broglie wavelength of an electron having kinetic energy \mathrm{E} is \lambda. If the kinetic energy of electron becomes \frac{E}{4}, then its de-Broglie wavelength will be :

Choose the correct answer: