The de Broglie wavelength of an electron with kinetic energy K_1 is λ_1 . If the kinetic energy increases to 5K_1 , the new wavelength is λ_2 = xλ_1 . The value of x is

Correct answer: C. 1 5 The de Broglie wavelength of an electron is given by λ = h 2mK For the initial state, λ_1 = h 2mK_1 For the final state, the kinetic energy is K_2 = 5K_1 . The new wavelength is: λ_2 = h 2m(5K_1) = 1 5 h 2mK_1 λ_2 = 1 5 λ_1 Comparing this with λ_2 = xλ_1 , we get x = 1 5 Answer: 1 5

MHT CET 202625 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 25 April
Shift: Evening Shift
Subject: Physics
Chapter: Dual Nature of Matter

MHT CET 2026 · 25 April · Evening Shift · Physics · Dual Nature of Matter

The de Broglie wavelength of an electron with kinetic energy K_1 is...

The de Broglie wavelength of an electron with kinetic energy

K_1

\lambda_1

. If the kinetic energy increases to

5K_1

, the new wavelength is

\lambda_2 = x\lambda_1

. The value of

x

Options

A $5$
B $\sqrt{5}$
C $\dfrac{1}{\sqrt{5}}$
D $\dfrac{1}{5}$

Correct answer

C. $\dfrac{1}{\sqrt{5}}$

Step-by-step solution

The de Broglie wavelength of an electron is given by

\lambda = \dfrac{h}{\sqrt{2mK}}

For the initial state,

\lambda_1 = \dfrac{h}{\sqrt{2mK_1}}

For the final state, the kinetic energy is

K_2 = 5K_1

. The new wavelength is:

\lambda_2 = \dfrac{h}{\sqrt{2m(5K_1)}} = \dfrac{1}{\sqrt{5}} \dfrac{h}{\sqrt{2mK_1}}

\lambda_2 = \dfrac{1}{\sqrt{5}} \lambda_1

Comparing this with

\lambda_2 = x\lambda_1

, we get

x = \dfrac{1}{\sqrt{5}}

Answer:

\dfrac{1}{\sqrt{5}}

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