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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