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MCQEasyJEE 2023de Broglie Relation

JEE Physics 2023 Question with Solution

An electron accelerated through a potential difference V1V_1 has a de-Broglie wavelength of λ\lambda. When the potential is changed to V2V_2, its de-Broglie wavelength increases by 50%50\%. The value of V1V2\frac{V_1}{V_2} is equal to :

  • A

    33

  • B

    94\frac{9}{4}

  • C

    32\frac{3}{2}

  • D

    44

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: An electron accelerated through potential difference V1V_1 has de-Broglie wavelength λ\lambda. When the potential becomes V2V_2, the wavelength becomes 1.5λ1.5\lambda.

Find: The value of V1V2\frac{V_1}{V_2}.

Using the relations:

KE=P22mKE=\frac{P^2}{2m}

and

P=hλP=\frac{h}{\lambda}

For potential difference V1V_1,

eV1=(hλ)22meV_1=\frac{\left(\frac{h}{\lambda}\right)^2}{2m}

For potential difference V2V_2, the wavelength increases by 50%50\%, so new wavelength is 1.5λ1.5\lambda.

eV2=(h1.5λ)22meV_2=\frac{\left(\frac{h}{1.5\lambda}\right)^2}{2m}

From the above two equations,

V1V2=(1.5)2\frac{V_1}{V_2}=(1.5)^2 V1V2=2.25=94\frac{V_1}{V_2}=2.25=\frac{9}{4}

Therefore, the correct option is B, that is 94\frac{9}{4}.

Wavelength-Potential Proportionality

Given: λ∝1V\lambda \propto \frac{1}{\sqrt{V}} for an electron accelerated through a potential difference.

Find: V1V2\frac{V_1}{V_2} when wavelength changes from λ\lambda to 1.5λ1.5\lambda.

Since

λ∝1V\lambda \propto \frac{1}{\sqrt{V}}

we get

λ2λ1=V1V2\frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}}

Now,

λ2λ1=1.5λλ=1.5\frac{\lambda_2}{\lambda_1}=\frac{1.5\lambda}{\lambda}=1.5

So,

V1V2=1.5\sqrt{\frac{V_1}{V_2}}=1.5 V1V2=(1.5)2=94\frac{V_1}{V_2}=(1.5)^2=\frac{9}{4}

Therefore, the correct option is B.

Common mistakes

  • Using λ∝V\lambda \propto \sqrt{V} instead of λ∝1V\lambda \propto \frac{1}{\sqrt{V}}. This reverses the dependence and gives the wrong ratio. Always remember that increasing accelerating potential decreases de-Broglie wavelength.

  • Treating a 50%50\% increase in wavelength as λ2=1.05λ\lambda_2=1.05\lambda or λ2=0.5λ\lambda_2=0.5\lambda. A 50%50\% increase means the new wavelength is 1.5λ1.5\lambda.

  • Comparing kinetic energies directly without substituting p=hλp=\frac{h}{\lambda}. The wavelength information enters through momentum, so first connect de-Broglie relation with kinetic energy before forming the voltage ratio.

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