Dual Nature of Matter

JEE Main level, numerical answer, medium

5 of 200

Question 5 of 200

The de Broglie wavelength for an electron accelerated through the potential difference of V1V_1 volt is λ1\lambda_1. When the potential difference is changed to V2V_2 volt, the associated de Broglie wavelength is increased by 50%50\%. If (V1/V2)=(9/α)(V_1/V_2) = (9/\alpha), then the value of α\alpha is __________.

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Answer: 4

The de Broglie wavelength of an electron accelerated through a potential difference VV is given by λ=h2meV\lambda = \dfrac{h}{\sqrt{2meV}}.

This implies λ∝1V\lambda \propto \dfrac{1}{\sqrt{V}}.

Given that the new wavelength λ2\lambda_2 is increased by 50%50\%, we have λ2=λ1+0.5λ1=32λ1\lambda_2 = \lambda_1 + 0.5\lambda_1 = \dfrac{3}{2}\lambda_1.

Taking the ratio of the wavelengths, we get λ1λ2=V2V1\dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{V_2}{V_1}}.

Substituting λ2=32λ1\lambda_2 = \dfrac{3}{2}\lambda_1, we get 23=V2V1\dfrac{2}{3} = \sqrt{\dfrac{V_2}{V_1}}.

Squaring both sides yields V2V1=49\dfrac{V_2}{V_1} = \dfrac{4}{9}, which gives V1V2=94\dfrac{V_1}{V_2} = \dfrac{9}{4}.

Comparing this with the given relation V1V2=9α\dfrac{V_1}{V_2} = \dfrac{9}{\alpha}, we find α=4\alpha = 4.

Answer: 44

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