Dual Nature of Matter

JEE Main level, single correct, medium

32 of 200

Question 32 of 200

An electron accelerated through a potential difference V1{V}_{1} has a de-Broglie wavelength of λ\lambda. When the potential is changed to V2{V}_{2}, its de-Broglie wavelength increases by 5050%. The value of (V1V2)\left(\dfrac{{V}_{1}}{{V}_{2}}\right) is equal to :

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Answer: (B) 94\dfrac{9}{4}

Let initial wavelength be λ\lambda, then after 5050% increase wavelength will become 1.5λ1.5\lambda.
Now, KE=\dfrac{{P}^{2}}{2m}=eV&P=\dfrac{h}{\lambda }
Therefore, eV=(hλ)22meV=\dfrac{{\left(\dfrac{h}{\lambda }\right)}^{2}}{2m}.
So we can write
eV1=(hλ)22m...Eq(1)e{V}_{1}=\dfrac{{\left(\dfrac{h}{\lambda }\right)}^{2}}{2m}...Eq(1) and
eV2=(h1.5λ)22m...Eq(2)e{V}_{2}=\dfrac{{\left(\dfrac{h}{1.5\lambda }\right)}^{2}}{2m}...Eq(2)
From both equations, we get
V1V2=(1.5)2=94\dfrac{{V}_{1}}{{V}_{2}}={\left(1.5\right)}^{2}=\dfrac{9}{4}

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