The de Broglie wavelength for an electron accelerated through the potential difference of V1 volt is lambda1. When the potential difference is changed to V2 volt, the associated de Broglie wavelength is increased by 50%. If (V1/V2) = (9/a), then the value of a is _______.

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q47 – JEE-MAIN 2026

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Home Physics Dual Nature of MatterQuestion #74

The de Broglie wavelength for an electron accelerated through the potential difference of

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

(V_{1} / V_{2}) = (9 / a)

, then the value of

a

is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: De Broglie Wavelength Concept

Recall the relationship between de Broglie wavelength and the accelerating potential difference for an electron.

🧠 Full Solution Path

Step 1: Relate de Broglie wavelength to potential difference✦ Active

The de Broglie wavelength $λ$ for an electron accelerated through a potential difference $V$ is given by $λ = \frac{h}{\sqrt{2 m e V}}$ . Since Planck's constant ( $h$ ), electron mass ( $m$ ), and electron charge ( $e$ ) are constants, we can establish the proportionality:

λ \propto \frac{1}{\sqrt{V}}

Step 2: Formulate the ratio of wavelengths and potentials○ Expand

Given that the de Broglie wavelength is increased by 50\% when the potential difference changes from $V_{1}$ to $V_{2}$ , the new wavelength $λ_{2}$ is $1.5$ times the initial wavelength $λ_{1}$ . Using the proportionality from Step 1, we can write:

\frac{λ_{2}}{λ_{1}} = \sqrt{\frac{V_{1}}{V_{2}}}

Substituting

λ_{2} = 1.5 λ_{1}

into the equation:

\frac{1.5 λ_{1}}{λ_{1}} = \sqrt{\frac{V_{1}}{V_{2}}} ⟹ 1.5 = \sqrt{\frac{V_{1}}{V_{2}}}

💡 Teacher's Secret Hint

Ensure correct handling of the percentage increase in wavelength.

Step 3: Calculate the value of

a

○ Expand

Square both sides of the equation from Step 2 to eliminate the square root:

(1.5)^{2} = \frac{V_{1}}{V_{2}} ⟹ 2.25 = \frac{V_{1}}{V_{2}}

We are given that

\frac{V_{1}}{V_{2}} = \frac{9}{a}

. Equating the two expressions for

\frac{V_{1}}{V_{2}}

2.25 = \frac{9}{a}

Solving for

a

a = \frac{9}{2.25} = \frac{9}{\frac{9}{4}} = 9 \times \frac{4}{9} = 4

💡 Teacher's Secret Hint

Double-check the arithmetic when converting decimals to fractions for simplification.

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Physics Question 47 – JEE-MAIN 2026

Clue 1 of 3: De Broglie Wavelength Concept

🧠 Full Solution Path