Dual Nature Of Matter Question 9
Question 9 - 30 January - Shift 2
An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases by $50 %$. The value of $(\frac{V_1}{V_2})$ is equal to :
(1) 3
(2) $\frac{9}{4}$
(3) $\frac{3}{2}$
(4) 4
Show Answer
Answer: (2)
Solution:
Formula: De Broglie wavelength
$KE=\frac{P^{2}}{2 m}, P=\frac{h}{\lambda}$
$eV_1=\frac{(\frac{h}{\lambda})^{2}}{2 m}$
$eV_2=\frac{(\frac{h}{1.5 \lambda})^{2}}{2 m}$
$\frac{V_1}{V_2}=(1.5)^{2}=\frac{9}{4}$
