Question

The de-Broglie wavelength associated with a material particle when it is accelerated through a potential difference of 150 volt is 1 Å. What will be the de-broglie wavelength associated with the same particle when it is accelerated through a potential difference of 4500 V?

विकल्प

• ${\displaystyle \frac{1}{\sqrt{3}}}$ Å

• ${\displaystyle \frac{1}{\sqrt{30}}}$Å

• ${\displaystyle \frac{1}{\sqrt{300}}}$ Å

• ${\displaystyle \frac{1}{3}}$ Å

Answer 1

${\displaystyle \frac{1}{\sqrt{30}}}$Å

Explanation:

As we know ${\displaystyle \lambda =\frac{\text{h}}{\sqrt{2\text{mK}}}}$, K = eV_o , K ∝ V_o

${\displaystyle \lambda \propto \frac{1}{\sqrt{\text{K}}}}$

${\displaystyle \lambda \propto \frac{1}{{\sqrt{\text{V}}}_{\text{O}}}}$

${\displaystyle \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{{\text{V}}_1}{{\text{V}}_2}}}$

${\displaystyle \Rightarrow {\lambda }_2=\sqrt{\frac{150}{4500}}}$Å

${\displaystyle =\frac{1}{\sqrt{30}}\AA }$