The de-Broglie wavelength for an electron accelerated through the potential difference volt is . When the potential

Question

The de-Broglie wavelength for an electron accelerated through the potential difference $V_{1}$ volt is $λ_{1}$ . When the potential difference is changed to $V_{2}$ volt, the associated de-Broglie wavelength is increased by 50%. If $(\frac{V_{1}}{V_{2}}) = (\frac{9}{α})$ , then the value of $α$ is ________. [2026]

Smart Achievers · Accepted Answer

(4)

$λ_{1} = \frac{h}{m v} = \frac{h}{\sqrt{2 m e V_{1}}} = λ$

$λ_{2} = \frac{h}{\sqrt{2 m e V_{2}}} = \frac{3 λ}{2} = \frac{3}{2} \frac{h}{\sqrt{2 m e V_{1}}}$

$V_{2} = \frac{4}{9} \cdot V_{1} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{9}{4} \Rightarrow α = 4$

Solution

Ans.
(4)

$λ_{1} = \frac{h}{m v} = \frac{h}{\sqrt{2 m e V_{1}}} = λ$

$λ_{2} = \frac{h}{\sqrt{2 m e V_{2}}} = \frac{3 λ}{2} = \frac{3}{2} \frac{h}{\sqrt{2 m e V_{1}}}$

$V_{2} = \frac{4}{9} \cdot V_{1} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{9}{4} \Rightarrow α = 4$

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Solution

Q. The de-Broglie wavelength for an electron accelerated through the potential difference V1 volt is λ1. When the potential difference is changed to V2 volt, the associated de-Broglie wavelength is increased by 50%. If (V1V2)=(9α), then the value of α is ________. [2026]

Ans. (4) λ1=hmv=h2meV1=λ λ2=h2meV2=3λ2=32h2meV1 V2=49·V1⇒V1V2=94⇒α=4

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Ans.
(4)

$λ_{1} = \frac{h}{m v} = \frac{h}{\sqrt{2 m e V_{1}}} = λ$

$λ_{2} = \frac{h}{\sqrt{2 m e V_{2}}} = \frac{3 λ}{2} = \frac{3}{2} \frac{h}{\sqrt{2 m e V_{1}}}$

$V_{2} = \frac{4}{9} \cdot V_{1} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{9}{4} \Rightarrow α = 4$