An electron accelerated through a potential difference V_1 has a … - Vidyadip

MCQ

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 $\left(\frac{ V _1}{ V _2}\right)$ is equal to :

A
$3$
✓
$\frac{9}{4}$
C
$\frac{3}{2}$
D
$4$

✓

Answer

Correct option: B.

$\frac{9}{4}$

b
$KE =\frac{ P ^2}{2 m }, \quad P =\frac{ h }{\lambda}$

$eV _1=\frac{\left(\frac{ h }{\lambda}\right)^2}{2 m }$

$eV _2=\frac{\left(\frac{ h }{1.5 \lambda}\right)^2}{2 m }$

$\frac{ V _1}{ V _2}=(1.5)^2=\frac{9}{4}$

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