An electron accelerated through a potential difference V1 has a de-Broglie wavelength of λ. When the potential is changed to V2, its de-Broglie wavelength increases by 50 %. The value of ((V1/V2)) is equal to

Question

An electron accelerated through a potential difference V1 has a de-Broglie wavelength of λ. When the potential is changed to V2, its de-Broglie wavelength increases by 50 %. The value of ((V1/V2)) is equal to (A) 4 (B) (9/4) (C) 3 (D) (3/2)

Tardigrade Admin · Accepted Answer

KE =( P 2/2 m ), P =( h /λ) eV 1=((( h /λ))2/2 m ) eV 2=((( h /1.5 λ))2/2 m ) ( V 1/ V 2)=(1.5)2=(9/4)

Q. An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $λ$ . 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

4

$\frac{9}{4}$

3

$\frac{3}{2}$

$K E = \frac{P ^{2}}{2 m},$
$P = \frac{h}{λ}$
$e V_{1} = \frac{( \frac{h}{λ} ) ^{2}}{2 m}$
$e V_{2} = \frac{( \frac{h}{1.5 λ} ) ^{2}}{2 m}$
$\frac{V _{1}}{V _{2}} = (1.5)^{2} = \frac{9}{4}$

Q. An electron accelerated through a potential difference V1​ has a de-Broglie wavelength of λ. When the potential is changed to V2​, its de-Broglie wavelength increases by 50%. The value of (V2​V1​​) is equal to

4

49​

3

23​

KE=2mP2​, P=λh​ eV1​=2m(λh​)2​ eV2​=2m(1.5λh​)2​ V2​V1​​=(1.5)2=49​

Q. An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $λ$ . 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

$\frac{9}{4}$

$\frac{3}{2}$

$K E = \frac{P ^{2}}{2 m},$
$P = \frac{h}{λ}$
$e V_{1} = \frac{( \frac{h}{λ} ) ^{2}}{2 m}$
$e V_{2} = \frac{( \frac{h}{1.5 λ} ) ^{2}}{2 m}$
$\frac{V _{1}}{V _{2}} = (1.5)^{2} = \frac{9}{4}$