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Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would
A
Increase by 4 times
B
Increase by 2 times
C
Decrease by 2 times
D
Decrease by 4 times
Detailed Solution
For an electron accelerated through a potential difference V, kinetic energy = eV and momentum $p = \sqrt{2meV}$.
de Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$, so $\lambda \propto \frac{1}{\sqrt{V}}$
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{25}{100}} = \frac{1}{2}$
$\lambda_2 = \frac{\lambda_1}{2}$
So the wavelength decreases by 2 times.
Note: the source prints '100 km'; it is 100 kV.
de Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$, so $\lambda \propto \frac{1}{\sqrt{V}}$
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{25}{100}} = \frac{1}{2}$
$\lambda_2 = \frac{\lambda_1}{2}$
So the wavelength decreases by 2 times.
Note: the source prints '100 km'; it is 100 kV.