Question 287: Dual Nature of Matter and Radiation

Question

An electron accelerated by a potential difference ' $V$ ' has de-Broglie wavelength ' $\lambda$ '. If the electron is accelerated by a potential difference ' 9 V ', its de-Broglie wavelength will be

Accepted Answer

Given:
An electron accelerated by potential difference $V$ has de-Broglie wavelength $\lambda$.
We have to find: The de-Broglie wavelength when it is accelerated by $9V$.
Step 1: Formula for de-Broglie wavelength of an electron
The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is:
$ \lambda = \frac{h}{p} $
When an electron is accelerated by $V$, its kinetic energy $K = eV$.
Momentum:
$ K = \frac{1}{2}mv^2 = eV \implies v = \sqrt{\frac{2eV}{m}} $
$ p = mv = m \sqrt{ \frac{2eV}{m} } = \sqrt{2meV} $
So,
$ \lambda = \frac{h}{\sqrt{2meV}} $
Step 2: Relationship between wavelength and potential
From the formula:
$ \lambda \propto \frac{1}{\sqrt{V}} $
Step 3: Compare for $V$ and $9V$
Let $\lambda_1$ be for $V$ and $\lambda_2$ for $9V$.
$ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V}{9V}} = \sqrt{\frac{1}{9}} = \frac{1}{3} $
$ \lambda_2 = \frac{\lambda_1}{3} $
Step 4: Conclusion
The answer is:
Option B
: $\frac{\lambda}{3}$

Answer

$\frac{\lambda}{4.5}$

Answer

$\frac{\lambda}{3}$

Answer

$\frac{\lambda}{2}$

Answer

$\lambda$

Dual Nature of Matter and Radiation

An electron accelerated by a potential difference ' V ' has de-Broglie wavelength ' \lambda '. If the electron is accelerated by a potential difference ' 9 V ', its de-Broglie wavelength will be

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