Question

The de Broglie wavelength of a particle of mass m and kinetic energy E is given by

• A. $\frac{h}{\sqrt{mE}}$
• B. $\frac{h}{\sqrt{2mE}}$
• C. $\frac{h}{2mE}$
• D. $\frac{h}{\sqrt{3mE}}$
• E. $\frac{h}{mE}$

Hint:

This formula shows that as a particle's energy increases, its wavelength decreases. For an electron accelerated through a potential $V$, $E = eV$, so $\lambda = \frac{h}{\sqrt{2meV}}$.

Answer 1

The Correct Option is B

Concept:
Physics - Dual Nature of Radiation and Matter.
Step 1: Recall the de Broglie wavelength formula.
The de Broglie wavelength ($\lambda$) is related to the momentum ($p$) of a particle: $$ \lambda = \frac{h}{p} $$
Step 2: Relate momentum to kinetic energy.
Kinetic energy ($E$) is given by $E = \frac{1}{2}mv^{2}$. Multiplying and dividing by $m$ gives: $$ E = \frac{m^{2}v^{2}}{2m} = \frac{p^{2}}{2m} $$ Solving for momentum ($p$): $$ p = \sqrt{2mE} $$
Step 3: Substitute momentum back into the wavelength formula.
$$ \lambda = \frac{h}{\sqrt{2mE}} $$