Question

Electron beam used in an electron microscope, when accelerated by a voltage of $20 kV$, has a de-Broglie wavelength of $\lambda_0$. If the voltage is increased to $40 kV$, then the de-Broglie wavelength associated with the electron beam would be:

• A. $\frac{\lambda_0}{2}$
• B. $3 \lambda_0$
• C. $\frac{\lambda_0}{\sqrt{2}}$
• D. $9 \lambda_0$

Answer 1

Answer: (C)

$\frac{\lambda_0}{\sqrt{2}}$

Answer 2

When electron is accelerated through potential difference V, then
K.E. =eV
⇒λ=2m(KE)​h​=2meV​h​
∴λαV​1​
∴λ0​λ​=4020​​
∴λ=2​λ0​​