De-Broglie wavelength associated with an electron accelerated through a potential difference ' $\mathrm{V}$…
De-Broglie wavelength associated with an electron accelerated through a potential difference ' $\mathrm{V}$ ' is ' $\lambda$ '. When the accelerating potential is increased to ' $4 \mathrm{~V}$ ', de-Broglie wavelength.
reduces to half
remains the same
reduces to $(1 / 4)^{\text {th }}$
increases by $25 \%$
Solution
De-Broglie wavelength, $\lambda=\frac{1.228}{\sqrt{V}}(\mathrm{~nm})$
If $\mathrm{V}$ is increased to $4 \mathrm{~V}$, then $\lambda$ will become $\frac{1}{\sqrt{4}}$ times or $\frac{1}{2}$ times.