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Problems · Problem 2.13

Q.The mass of an electron is 9.1×10−31 kg9.1 \times 10^{-31}\ kg. If its K.E. is 3.0×10−25 J3.0 \times 10^{-25}\ J, calculate its wavelength.

Rajasthan RbseTextbookSubjective· 2mImportance★★★★★
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The de Broglie wavelength of an electron is found by relating its kinetic energy to momentum, then using λ=h/p\lambda = h/p. For the given values, the wavelength comes out to be approximately 8.96×10−7 m8.96 \times 10^{-7}\ \text{m} (or 896 nm).

The key idea here is that every moving particle has a wavelength associated with it — that's the de Broglie hypothesis. For an electron, even though we usually think of it as a particle, when it's moving it also behaves like a wave. The wavelength depends on its momentum, not directly on its energy. So when you're given kinetic energy, you first need to find momentum.

Let's work through it step by step.

  1. Recall the de Broglie relation The wavelength λ\lambda of a particle is given by:

λ=hp\lambda = \frac{h}{p}

where h=6.63×10−34 J⋅sh = 6.63 \times 10^{-34}\ \text{J·s} is Planck's constant, and pp is the linear momentum of the particle. This is the fundamental formula — everything else is just algebra to find pp.

  1. Relate kinetic energy to momentum For a non-relativistic particle (which an electron at this energy certainly is — we'll check later), kinetic energy KK and momentum pp are connected by:

K=p22mK = \frac{p^2}{2m}

So we can solve for pp:

p=2mKp = \sqrt{2mK}

Tip

Always check if the electron is relativistic. A quick rule: if KK is much smaller than the electron's rest energy (mec2≈8.2×10−14 Jm_e c^2 \approx 8.2 \times 10^{-14}\ \text{J}), you're safe using the classical formula. Here K=3.0×10−25 JK = 3.0 \times 10^{-25}\ \text{J} — that's about 10 orders of magnitude smaller, so non-relativistic is perfectly fine.

  1. Plug in the numbers

    Given:

    • m=9.1×10−31 kgm = 9.1 \times 10^{-31}\ \text{kg}
    • K=3.0×10−25 JK = 3.0 \times 10^{-25}\ \text{J}

    First compute 2mK2mK:

2mK=2×(9.1×10−31)×(3.0×10−25)2mK = 2 \times (9.1 \times 10^{-31}) \times (3.0 \times 10^{-25})

=2×9.1×3.0×10−56= 2 \times 9.1 \times 3.0 \times 10^{-56}

=54.6×10−56=5.46×10−55= 54.6 \times 10^{-56} = 5.46 \times 10^{-55}

So:

p=5.46×10−55p = \sqrt{5.46 \times 10^{-55}}

Take the square root:

5.46≈2.336\sqrt{5.46} \approx 2.336

10−55=10−27.5=10−27×10−0.5=10−27×110≈10−27×0.3162\sqrt{10^{-55}} = 10^{-27.5} = 10^{-27} \times 10^{-0.5} = 10^{-27} \times \frac{1}{\sqrt{10}} \approx 10^{-27} \times 0.3162

So:

p≈2.336×0.3162×10−27p \approx 2.336 \times 0.3162 \times 10^{-27}

≈0.7386×10−27=7.386×10−28 kg⋅m/s\approx 0.7386 \times 10^{-27} = 7.386 \times 10^{-28}\ \text{kg·m/s} …

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