Question

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. 4
• B. $\frac{9}{4}$
• C. 3
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$\frac{9}{4}$

Answer 2

$KE=\frac {P^2}{2m}$

$P=\frac hλ$

$eV_1​=\frac {(\frac hλ​)^2​}{2m}$ ……… (1)

$eV_2​=\frac {(\frac {h}{1.5λ}​)^2​}{2m}$ ……….. (2)

From eq (1) and eq (2),
$\frac {V_1}{V_2}= (1.5)^2$

$\frac {V_1}{V_2}=2.25$

$\frac {V_1}{V_2}=\frac 94$

So, the correct option is (B): $\frac 94$.